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Electric Charges and Fields Class 12 Important Questions: CBSE Physics Chapter 1

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Class 12 Physics is a crucial subject for students preparing for board exams and competitive tests. The chapter "Electric Charges and Fields" forms the foundation of electromagnetism, covering essential concepts such as Coulomb's law, electric field, and dipole moments. Practicing important questions from this chapter ensures a thorough understanding and enhances problem-solving skills. Accessing a FREE PDF with these questions makes preparation accessible and organized for students aiming to score high.


Download the Class 12 Physics  Syllabus for detailed coverage and explore our exclusive Class 12 Physics Important Questions PDF to solidify your understanding and excel in your exams.

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Access Class 12 Physics Chapter 1: Electric Charges and Fields Important Questions

Very Short Answer Questions – 1 Marks

1. Does the Force Between Two Point Charges Change if the Dielectric Constant of the Medium in Which they are Kept is Increased?

Ans:  Dielectric constant of a medium is given by 

k=FVFM=force between the charges in vaccumforce between two charges in medium

FM=FVk

From the above expression, it is clear that as k is increased, FM gets decreased.


2. A Charged Rod P Attracts a Rod R Whereas P Repels another Charged Rod Q. What Type of Force is Developed Between  Q and R ?

Ans: Suppose that the rod P is negatively charged. As it attracts rodR, it can be said that R is positively charged. 

Also, since P repels rod Q, it can be said that Q is negatively charged. 

Clearly, the force between Q and R would be attractive in nature.


3. Which Physical Quantity has its S.I. Unit 

1. Cm ?

2. N/C ?

Ans:

1. The S.I. unit of electric dipole moment is Cm

2. The S.I. unit of electric field intensity is N/C.


4. Define One Coulomb.

Ans: Charge on a body is said to be 1 coulomb if it experiences a force of repulsion or attraction of 9×109N from another equal charge when they are separated by a distance of 1m.


Short Answer Questions in Electric Charges and Fields Important Questions with Answers PDF – 2 Marks

1. A Free Proton and a Free Electron are Placed in a Uniform Field. Which of the Two Experiences Greater Force and Greater Acceleration?


Free proton


Ans: Force on both the electron as well as the proton in the uniform field would be equal because F=kq and it is known that charge on both electron and proton are the same. On the other hand, since acceleration is given by a=Fm and as the mass of a proton is more than that of an electron, the acceleration of the electron would be more.


2. No Two Electric Lines of Force Can Intersect Each Other. Why?          

Ans:  Two electric lines of force can never intersect each other.  Suppose if they intersect, then at the point of intersection, there can be two tangents drawn.

These two tangents are supposed to represent two directions of electric field lines, which is not possible at a particular point.


3. The Graph Shows the Variation of Voltage V Across the Plates of Two Capacitors A and B Versus Increase of Charge Q Stored on Them. Which of the Two Capacitors Have Higher Capacitance? Give Reason for Your Answer.


voltage graph


Ans: 

It is known that C=QV

Clearly, for a given charge Q,

C1V

Now, from the given graph, it is seen that VA<VB.

Therefore, it can be concluded that CA>CB.


4. An Electric Dipole When Held at 30 with Respect to a Uniform Electric Field of 104N/C Experiences a Torque of 9×1026Nm. Calculate Dipole Moment of the Dipole?

Ans: 

It is given that

θ = 30⁰

τ = 9 x 10⁻²⁶Nm

E = 10⁴N/C

Dipole moment P needs to be calculated.

It is known that torque is given by τ=PEsinθ.

Clearly,

P=τEsinθ

P=9×1026104×sin30=9×1026×104×21

P=18×1030Cm


5. 

a) Explain the Meaning of the Statement 'Electric Charge of a Body is Quantized’.

Ans:  The statement ‘electric charge of a body is quantized’ suggests that only integral (1, 2,3,4,...,n) number of electrons can be transferred from one body to another. 

This further suggests that charges are not transferred in fractions. 

Hence, a body possesses its total charge only in integral multiples of electric charges.


b) Why Can One Ignore the Quantization of Electric Charge When Dealing With Macroscopic I.e., Large Scale Charge?

Ans: When dealing with macroscopic or large-scale charges, the charges used are huge in number as compared to the magnitude of electric charge. 

Hence, the quantization of electric charge is of no use on a macroscopic scale. 

Therefore, it is ignored and considered that electric charge is continuous.


6. When a Glass Rod is Rubbed with a Silk Cloth, Charges Appear on Both. A Similar Phenomenon is Observed with Many Other Pairs of Bodies. Explain How This Observation is Consistent With the Law of Conservation of Charge

Ans: Rubbing is a phenomenon in which there is production of charges equal in magnitude but opposite in nature on the two bodies which are rubbed with each other. 

It is also seen that during such a phenomenon, charges are created in pairs. This phenomenon of charging is called as charging by friction. 

The net charge on a system of two rubbed bodies is equal to zero. This is because equal number of opposite charges in both the bodies annihilate each other. 

Clearly, when a glass rod is rubbed with a silk cloth, opposite natured charges appear on both these bodies. 

Thus, this phenomenon is consistent with the law of conservation of energy. As already mentioned, a similar phenomenon is observed with many other pairs of bodies too.


7. 

a) An Electric Field Line is a Continuous Curve. That is, a Field Line Cannot Have Sudden Breaks. Why Not?

Ans: An electrostatic field line is a continuous curve as it is known that a charge experiences a continuous force when traced in an electrostatic field. 

Also, the field line cannot have sudden breaks because the charge moves continuously and does not have the potential to jump from one point to another.


b) Explain Why Two Field Lines Never Cross Each Other at Any Point? 

Ans: Suppose two field lines cross each other at a particular point, then electric field intensity will show two directions at that point of intersection. 

This is impossible. Thus, two field lines can never cross each other.


8. An Electric Dipole with Dipole Moment 4×109Cm is Aligned at 30 with Direction of a Uniform Electric Field of Magnitude 5×104NC1. Calculate the Magnitude of the Torque Acting on the Dipole.

Ans: 

It is given that:

Electric dipole moment, p=4×109Cm 

Angle made by p with uniform electric field, θ=30 

Electric field, E=5×104NC1 

Torque acting on the dipole is given by τ=pEsinθ.

τ=4×109×5×104×sin30

τ=20×105×12

τ=104Nm 

Thus, the magnitude of the torque acting on the dipole is 104Nm.


9. Figure Below Shows Tracks of Three Charged Particles in a Uniform Electrostatic Field. Give the Signs of the Three Charges. Which Particle Has the Highest Charge to Mass Ratio?


Electrostatic Field


Ans: Since unlike charges attract and like charges repel each other, the particles 1 and 2 moving towards the positively charged plate are negatively charged whereas the particle 3 that moves towards the negatively charged plate is positively charged.

Since the charge to mass ratio is directly proportional to the amount of deflection for a given velocity, particle 3 would have the highest charge to mass ratio.


10. What is the Net Flux of the Uniform Electric Field of Exercise 1.15 through a Cube of Side 20cm Oriented So That Its Faces Are Parallel to the Coordinate Planes?

Ans: It is given that all the faces of the cube are parallel to the coordinate planes. 

Clearly, the number of field lines entering the cube is equal to the number of field lines entering out of the cube. 

As a result, the net flux through the cube can be calculated to be zero.


11. Careful Measurement of the Electric Field at the Surface of a Black Box Indicate That the Net Outward Flux Through the Surface of the Box is 8.0×103Nm2/C.

a) What is the Net Charge Inside the Box?

Ans:

It is given that,

The net outward flux through surface of the box, ϕ=8.0×103Nm2/C.

For a body containing net charge q, flux is given by ϕ=qε0

Where,

ε0= Permittivity of free space =8.854×1012N1C2m2

Therefore, the charge q is given by  q=ϕε0.

q=8.854×1012×8.0×103

q=7.08×108

q=0.07μC

Therefore, the net charge inside the box is 0.07μC


b) If the Net Outward Flux Through the Surface of Box Were Zero, Could You Conclude That There Were No Charges Inside the Box? Why or Why Not?

Ans: No. The net flux entering out through a body depends on the net charge contained in the body. If the net flux is given to be zero, then it can be inferred that the net charge inside the body is zero. 

For the net charge associated with a body to be zero, the body can have equal amounts of positive and negative charges and thus, it is not necessary that there were no charges inside the box.


Short Answer Questions – 3 Marks

1. A Particle of Mass m and Charge q is Released from Rest in a Uniform Electric Field of Intensity E. Calculate the Kinetic Energy Attained by this Particle after Moving a Distance Between the Plates.                                                                                                            

Ans: 

We have the electrostatic force on a charge q in electric field E given by, 

F=qE.........................................(1) 

Also, we have Newton’s second law of motion given by, 

F=ma.........................................(2)

From (1) and (2), 

a=qEm……………………………………………… (3)

We have the third equation of motion given by,

v2u2=2as

Since the charged particle is initially at rest,

u=0

v2=2as…………………………………… (4)

We have the expression for kinetic energy given by, 

KE=12mv2……………………………………. (5)

Substituting (4) in (5) we get, 

KE=12m(2as)=mas…………………………….. (6)

Substituting (3) in (6) to get, 

KE=m×(qEm)×s

Therefore, we have the kinetic energy obtained by the particle of charge q on moving a distance s in electric field E given by, 

KE=qEs


2. Two Charges +q and +9q are Separated by a Distance of 10a. Find The Point on the Line Joining the Two Charges Where Electric Field is Zero. 


Two Charges


Ans: 

Let P be the point (at x distance from charge +q) on the line joining the given two charges where the electric field is zero. 

We know that the electric field at a point at r distance from any charge q is given by, 

E=Kqr2

Electric field due to charge +q at point P would be, 

E1=K(+q)x2……………………………………………… (1)

Electric field due to charge +9q at point P would be, 

E2=K(+9q)(10ax)2………………………………………….. (2)

Since the net electric field at point P is zero,

E1+E2=0

|E1|=|E2|

From (1) and (2),

K(+q)x2=K(+9q)(10ax)2

(10ax)2=9x2

10x=3x

10a=4x

x=104a=2.5a

Therefore, we found the point on the line joining the given two charges where the net electric field is zero to be at a distance x=2.5a from charge q and at a distance 10ax=10a2.5a=7.5a from charge 9q.


3. 

a) Define the Term Dipole Moment P of an Electric Dipole Indicating its Direction and also Give Its S.i. Unit. 

Ans:

Electric dipole moment is defined as the product of the magnitude of either of the two charges of the dipole and their distance of separation which would be the length of dipole. Mathematically,

P=2lq……………………………………………………. (1)

where 2lis the length of the dipole and q is the charge. 

The direction of dipole is from ve to +ve charge and its S.I. unit is coulomb meter(Cm)


b) An Electric Dipole is Placed in a Uniform Electric Field E. Deduce the Expression for the Torque Acting on It.

Ans. 

Consider a dipole placed in uniform electric field E making an angle θ with it. 


making an angle


Now, we know that the force acting on the given dipole will be the electrostatic force and this will be the cause for the resultant force. We have the expression for torque given by, 

τ=F×x…………………………………….. (2)

Where, F is the force on the dipole and x is the perpendicular distance. 

Where, force F is given by, 

F=qE…………………………………………….. (3) 

From the figure we have, 

sinθ=BNAB

BN=ABsinθ=2lsinθ

But BNhere is the perpendicular distance x, so, equation (2) becomes,

τ=qE×2lsinθ=(2lq)Esinθ

But from (1), P=2lq

Now, we could give the torque on the dipole as, 

τ=PEsinθ=P×E


4. A Sphere S1 of Radius R1 Encloses a Charge Q. If there is another Concentric SphereS2 of Radius R2(R2>R1)  and there is no Additional Change Between S1 and S2, then Find the Ratio of Electric Flux through S1  and S2.


Concentric Sphere


Ans: 

We may recall that the expression for electric flux through a surface enclosing charge q by Gauss’s law is given by, 

ϕ=qε0

Where, ε0is the permittivity of the medium. 

Now the electric flux through sphere S1 is given by, 

ϕS1=Qε0 …………………………………….. (1)

Since there is no additional charge between the given two spheres, the flux through sphere S2 is given by, 

ϕS2=Qε0…………………………………….. (2)

We could now get the ratio of flux through spheres S1 and S2,

ϕS1ϕS2=Qε0Qε0

ϕS1ϕS2=11

Therefore, we find the required ratio to be 1:1.


5. Electric Charge is Uniformly Distributed on the Surface of a Spherical Balloon. Show How Electric Intensity and Electric Potential Vary 

a) on the surface 

Ans:


Electric Charge


Electric field intensity on the surface of the balloon would be,

E=σε0 

Electric potential on the surface of the balloon would be, 

 V=KqR


b) inside 

Ans:


surface of the balloon


Electric field intensity inside the balloon would be,

E=0 

Electric potential inside the balloon would be, 

V=KqR


c) outside.

Ans:


outside balloon


Electric field intensity outside the balloon would be,

E=σε0R2r2 

Electric potential outside the balloon would be

V=Kqr

We could represent this variation graphically as, 

For electric field, 


graph 1


For electric potential, 


graph 2



6. Two Point Electric Charges q and 2q are Kept at a Distance d Apart from Each Other in Air. a Third Charge Q is to be Kept Along the Same Line in Such a Way That the Net Force Acting On q and 2q is Zero. Calculate the Position of Charge Q in Terms of q and d.

Ans: 


two point charge


For the net force on charge q and 2q to be zero, the third charge should be negative since other two given charges are positive.  

The force between two charges q1 and q2 separated by a distancer is given by Coulomb’s law as, 

F=Kq1q2r2

Force acting on charge Q due to q=14πε0qQx2

Force acting on charge Qdue to 2q=14πε02qQ(dx)2

Now for the given system to be in equilibrium,

KQqx2=K2Qq(dx)2……………………………………………….. (1)

From equations (1) and (2) we get, 

1x2=2(dx)2

2x2=(dx)2

2x=dx

x=d2+1

So, we found that the new charge Q should be kept between the given two charges at a distance of x=d2+1 from charge q.


7. What is the Force Between Two Small Charged Spheres Having Charges of 2×107C and 3×107C placed 30cm Apart in Air? 

Ans: 

We are given:

Charge of the first sphere, q1=2×107C

Charge of the second sphere, q2=3×107C

Distance between the two spheres, r=30cm=0.3m

Electrostatic force between the spheres is given by Coulomb’s law as,

F=q1q24πε0r2 ……………………………… (1)

Where, ε0=Permittivity of free space and,14πε0=9×109Nm2C2

Substituting the given values in (1), we get, 

F=9×109×2×107×3×107(0.3)2

F=6×103N

Hence, force between the two small charged spheres is found to have a magnitude of6×103N

Since both the given charges are positive, the resultant force would be repulsive as like charges repel each other. 


8. The Electrostatic Force on a Small Sphere of Charge 0.4μC Due to another Small Sphere of Charge 0.8μC in Air is 0.2N.

a) What is the Distance Between the Two Spheres? 

Ans:

It is given that:

Electrostatic force on the first sphere,F=0.2N 

Charge on the first sphere, q1=0.4μC=0.4×106C

Charge on the second sphere,q2=0.8μC=0.8×106C 

Electrostatic force between the spheres could be given by Coulomb’s law as,

F=q1q24πε0r2 ……………………………… (1)

Where, ε0=Permittivity of free space and,14πε0=9×109Nm2C2 

Rearranging (1) we get, 

r2=q1q24πε0F

Substituting the given values, 

r2=0.4×106×0.8×106×9×1090.2

r2=144×104

r=144×104

r=0.12m

Therefore, we found the distance between the given two spheres to be 0.2m.


b) What is the Force on the Second Sphere Due to the First?

Ans: Since, both the spheres attract each other with the same force, the force on the second sphere due to the first would be 0.2N


9. A Polythene Piece Rubbed With Wool is Found to Have a Negative Charge Of 3×107C.

a) Estimate the Number of Electrons Transferred (from Which to Which?)

Ans:

When polythene is rubbed against wool, certain number of electrons get transferred from wool to polythene. 

Hence, wool becomes positively charged on loosing electrons and polythene becomes negatively charged on gaining them.

Charge on the polythene piece, 

q=3×107C 

Charge of an electron, 

e=1.6×1019C 

Let the number of electrons transferred from wool to polythene be n, then, from the property of quantization of charge we have, 

q=ne 

n=qe

Now, on substituting the given values, we get, 

n=3×1071.6×1019

n=1.87×1012

Therefore, the number of electrons transferred from wool to polythene is found to be1.87×1012.


b) Is There a Transfer of Mass from Wool to Polythene?

Ans: Yes, during the transfer of electrons from wool to polythene, along with charge, mass is also transferred. 

Let m be the mass being transferred in the given case and me be the mass of the electron, then,

m=me×n 

m=9.1×1031×1.85×1012

m=1.706×1018kg

Hence, we found that a negligible amount of mass does get transferred from wool to polythene.


10. Consider a Uniform Electric Field E=3×103i^N/C

a) What is the Flux of This Field Through a Square of Side 10cm Whose Plane is Parallel to the Y-Z Plane? 

Ans:

It is given that:

Electric field intensity,E=3×103i^N/C 

Magnitude of electric field intensity,|E|=3×103N/C 

Side of the square,a=10cm=0.1m 

Area of the square, A=a2=0.01m2 

Since the plane of the square is parallel to the y-z plane, the normal to its plane would be directed in the x direction. So, angle between normal to the plane and the electric field would be,θ=0 

We know that the flux through a surface is given by the relation, 

ϕ=EAcosθ

Substituting the given values, we get, 

ϕ=3×103×0.01×cos0

ϕ=30Nm2/C

Now, we found the net flux through the given surface to be, ϕ=30Nm2/C.


b) What is the Flux Through the Same Square if the Normal to Its Plane Makes 60angle with the x-axis? 

Ans: 

When the plane makes an angle of 60 with the x-axis, the flux through the given surface would be,

ϕ=EAcosθ

ϕ=3×103×0.01×cos60

ϕ=30×12

ϕ=15Nm2/C

So, we found the flux in this case to be,ϕ=15Nm2/C. 


11. A Point Charge +10μC  is a Distance 5cm Directly Above the Center of a Square of Side 10cm, as Shown in Fig. 1.34. What is the Magnitude of the Electric Flux Through the Square? (Hint: Think of the Square as One Face of a Cube With Edge 10cm

Ans: 

Considering square as one face of a cube of edge 10cm with a charge q at its center, according to Gauss's theorem for a cube, total electric flux is through all its six faces.


Electric Flux


ϕtotal=qε0

The electric flux through one face of the cube could be given by, ϕ=ϕtotal6

ϕ=16qε0 

Permittivity of free space, ε0=8.854×1012N1C2m2.

The net charge enclosed, q=10μC=10×106C.

Substituting the values given in the question, we get, 

ϕ=16×10×1068.854×1012

ϕ=1.88×105Nm2C1

Therefore, electric flux through the square is found to be 1.88×105Nm2C1.


12. A Point Charge of 2.0μC is Kept at the Center of a Cubic Gaussian Surface of Edge Length 9cm. What is the Net Electric Flux Through the Surface? 

Ans: 

Let us consider one of the faces of the cubical Gaussian surface considered, which would be a square. 

Since, a cube has six such square faces in total, we could say that the flux through one surface would be one-sixth the total flux through the gaussian surface considered. 


Gaussian Surface


The net flux through the cubical Gaussian surface by Gauss’s law is given by, 

ϕtotal=qε0

So, the electric flux through one face of the cube would be, 

ϕ=ϕtotal6

ϕ=16qε0……………………………….. (1)

But we have, permittivity of free space, ε0=8.854×1012N1C2m2

Charge enclosed, q=10μC=10×106C.

Substituting the given values in (1) we get, 

ϕ=16×10×1068.854×1012

ϕ=1.88×105Nm2C1

Therefore, electric flux through the square surface is 1.88×105Nm2C1.


13. A Point Charge Causes an Electric Flux of 1.0×103Nm2/C to Pass Through a Spherical Gaussian Surface Of 10cm Radius Centered on the Charge. 

a) If the Radius of the Gaussian Surface Were Doubled, How Much Flux Would Pass Through the Surface? 

Ans:

Electric flux due to the given point charge, 

ϕ=1.0×103Nm2/C 

Radius of the Gaussian surface enclosing the point charge,

r=10.0cm 

Electric flux piercing out through a surface depends on the net charge enclosed by the surface from Gauss’s law.

It is independent of the dimensions of the arbitrary surface assumed to enclose this charge. 

If the radius of the Gaussian surface is doubled, then the flux passing through the surface remains the same i.e.,103Nm2/C.


b) What is the Value of the Point Charge?

Ans:  

Electric flux is given by the relation,

ϕtotal=qε0

Where,

q= net charge enclosed by the spherical surface 

Permittivity of free space, ε0=8.854×1012N1C2m2 

q=ϕε0

Substituting the given values,

q=1.0×103×8.854×1012=8.854×109C

q=8.854nC

Therefore, the value of the point charge is found to be 8.854nC.


14. A Conducting Sphere of Radius 10cmhas an Unknown Charge. If the Electric Field at a Point 20cm from the Center of the Sphere of Magnitude 1.5×103N/C is Directed Radially Inward, What Is the Net Charge on the Sphere?

Ans: 

We have the relation for electric field intensityE at a distance d from the center of a sphere containing net charge qis given by,

E=q4πε0d2 ……………………………………………… (1)

Where, 

Net charge, 

q=1.5×103N/C

Distance from the center, 

d=20cm=0.2m 

Permittivity of free space,

ε0=8.854×1012N1C2m2 

14πε0=9×109Nm2C2

From (1), the unknown charge would be, 

q=E(4πε0)d2

Substituting the given values we get, 

q=1.5×103×(0.2)29×109=6.67×109C

q=6.67nC

Therefore, the net charge on the sphere is found to be 6.67nC.


15. A Uniformly Charged Conducting Sphere of 2.4m Diameter Has a Surface Charge Density of 80.0μC/m2.


Uniformly Charged


a) Find the Charge on the Sphere. 

Ans:

Diameter of the sphere, 

d=2.4m 

Radius of the sphere,

r=1.2m 

Surface charge density, 

σ=80.0μC/m2=80×106C/m2 

Total charge on the surface of the sphere,

Q=Charge density × Surface area 

Q=σ×4πr2=80×106×4×3.14×(1.2)2

Q=1.447×103C

Therefore, the charge on the sphere is found to be 1.447×103C.


b) What is the Total Electric Flux Leaving the Surface of the Sphere?

Ans: 

Total electric flux (ϕtotal) leaving out the surface containing net charge Q is given by Gauss’s law as, 

ϕtotal=Qε0…………………………………………………. (1)

Where, permittivity of free space,

ε0=8.854×1012N1C2m2

We found the charge on the sphere to be, 

Q=1.447×103C

Substituting these in (1), we get, 

ϕtotal=1.447×1038.854×1012

ϕtotal=1.63×108NC1m2

Therefore, the total electric flux leaving the surface of the sphere is found to be 1.63×108NC1m2.


16. An Infinite Line Charge Produces a Field of Magnitude 9×104N/C at a Distance of 2cm. Calculate the Linear Charge Density.

Ans: 

Electric field produced by the given infinite line charge at a distance dhaving linear charge densityλ could be given by the relation,

E=λ2πε0d 

λ=2πε0Ed…………………………………….. (1)

We are given:

d=2cm=0.02m  

E=9×104N/C 

Permittivity of free space,

ε0=8.854×1012N1C2m2 

Substituting these values in (1) we get, 

λ=2π(8.854×1012)(9×104)(0.02)

λ=10×108C/m

Therefore, we found the linear charge density to be 10×108C/m.


17. Which Among the Curves Shown in Fig. 1.35 Cannot Possibly Represent Electrostatic Field Lines?

a) 


Field Lines


Ans: 

The field lines showed in (a) do not represent electrostatic field lines because field lines must be normal to the surface of the conductor which is a characterizing property of electric field lines.


b) 


surface of the conductor


Ans: 

The lines showed in (b) do not represent electrostatic field lines because field lines cannot emerge from a negative charge and cannot terminate at a positive charge since the direction of the electric field is from positive to negative charge. 


c) 


direction of the electric field


Ans:

The field lines showed in (c) do represent electrostatic field lines as they are directed outwards from positive charge in accordance with the property of electric field. 


d)


electrostatic field lines


Ans: 

The field lines showed in (d) do not represent electrostatic field lines because electric field lines should not intersect each other.


e) 


field lines


Ans: 

The field lines showed in (e) do not represent electrostatic field lines because electric field lines do not form closed loops


18. Suppose that the Particle in Exercise in 1.33 Is an Electron Projected With Velocity vx=2.0×106ms1. If E Between the Plates Separated by 0.5cm is 9.1×102N/C, Where Will the Electron Strike the Upper Plate? (|e|=1.6×1019C,me=9.1×1031kg )

Ans: 

We are given the velocity of the particle, vx=2.0×106ms1

Separation between the two plates, d=0.5cm=0.005m 

Electric field between the two plates, E=9.1×102N/C 

Charge on an electron, e=1.6×1019C 

mass of an electron, me=9.1×1031kg 

Letsbe the deflection when the electron strikes the upper plate at the end of the plate L, then, we have the deflection given by, 

s=qEL22mvx 

L=2dmvxqE

Substituting the given values, 

L=2×0.005×9.1×1031×(2.0×106)21.6×1019×9.1×102=0.025×102=2.5×104

L=1.6×102=1.6cm

Therefore, we found that the electron will strike the upper plate after travelling a distance of 1.6cm.


Short Answer Questions – 5 Marks

1.

a) The Expression of Electric Field Edue to a Point Charge at Any Point Near to it Is Defined By E=limq0Fq Where q  is the Test Charge and F is the force acting on it. What is the Significance of limq0 in this Expression?

Ans: The significance of  limq0 is that the test charge should be vanishingly small so that it is not disrupting the presence of the source charge.


b) Two Charges Each of Magnitude 2×107C but Opposite in Sign Forms a System. These Charges Are Located at Points A (0,0,10) and B (0,0,+10) Respectively. Distances Are Given in Cm. What Is the Total Charge and Electric Dipole Moment of the System?

Ans.: 

Total charge of the system =(+2×107)+(2×107)=0.

Electric dipole moment is:

P=q×2l

P=2×107×20×102

P=4×108cm

Also, the direction of electric dipole moment is along the negative z-axis.


2. 

A) Sketch Electric Lines of Force Due To 

i) Isolated Positive Charge (i.e., q>0) and 

ii) Isolated Negative Charge (i.e., q<0).

Ans. 

The sketch of isolated positive charge and isolated negative charge are as follows:


isolated positive charge


b) Two-point Charges q and q are Placed at a Distance of 2a apart. Calculate the Electric Field at a Point P Situated at a Distance r Along the Perpendicular Bisector of the Line Joining the Charges. What Is the Electric Field When r >> a?

Ans. 

As we know,

|E+q|=kqr2+a2

|Eq|=kqr2+a2

Since, |E+q|=|Eq|


Line Joining the Charges


|Enet|=E+q2+Eq2+2E+qEqcos2θ

|Enet|=2E+q2+2E+q2cos2θ

|Enet|=2E+q2(1+cos2θ)

|Enet|=2E+q2(2cos2θ)

|Enet|=4E+q2(cos2θ)

|Enet|=2E+qcosθ

|Enet|=2E+qar2+a2

|Enet|=2kqr2+a2ar2+a2

|Enet|=2akq(r2+a2)32

|Enet|=kP(r2+a2)32

For, r>>a (a can be neglected)

Therefore, we get,

|Enet|=kPr3


3. 

a) What Is an Equi-Potential Surface? Show That the Electric Field Is Always Directed Perpendicular to an Equi-Potential Surface.

Ans. 

An equipotential surface is a surface that has the same potential throughout.

As we know,

dW=Fdx

dW=(qE)dx

(Force on the test charge, F=(qE))

Since work done is moving a test charge along an equipotential surface is always zero,

0=(qE)dx

Edx=0

Edx


b) Derive an Expression for the Potential at a Point Along the Axial Line of a Short Electric Dipole

Ans.

Consider an electric dipole of dipole length 2a and point P on the axial line such that OP=r, where O is the centre of the dipole.


electric dipole


Electric potential at point P due to the dipole is given by:

V=VPA+VPB

V=K(q)(r+a)+K(+q)(ra)

V=Kq[1ra1r+a]

V=Kq[r+ar+a(ra)(r+a)]

V=Kq2ar2a2       (P=2aq)

V=KPr2a2

For a dipole having short length, a can be neglected.

This gives,

V=KPr2


4. Check if the Ratio Ke2Gmemp is Dimensionless. Look up at the Table of Physical Constants and Determine the Value of This Ratio. What Does This Ratio Signify?

Ans. 

The given ratio is Ke2Gmemp.

Where, G is Gravitational constant. Its unit is Nm2kg2.

me and mp are the masses of electron and proton respectively. Their unit is kg.

e is the electric charge. Its unit is C.

εo is the permittivity of free space. Its unit is Nm2C2.

Therefore, the unit of the given ratio Ke2Gmemp is

=[Nm2C2][C2][Nm2kg2][kg][kg]

And its dimensions can be related to =[M0L0T0]

Hence, the given ratio is dimensionless.

We know,

e=1.6×1019C

G=6.67×1011Nm2kg2

me=9.1×1031kg

mp=1.66×1027kg

Hence, the numerical value of the given ratio is

Ke2Gmemp=9×109×(1.6×1019)26.67×1011×9.1×1031×1.66×10272.3×1039

This ratio is showing the ratio of electric force to the gravitational force between a proton and an electron, keeping distance between them constant.


5. Four-point Charges qA=2μC, qB=5μC, qC=2μC, qD=5μC are Located at the Corners of a Square ABCD of Side 10 cm. What is the Force on a Charge of 1μC Placed at the Centre of the Square?

Ans. 

In the given figure, there is a square having length of each side is 10cm and four charges placed at its corners. O is the centre of the square.


centre of the square


AB, BC, CD and AD are the sides of the square. Each of length is 10cm

AC and BD are the diagonals of the square of length 102cm.

AO, OB, OC, OD are of length 52cm.

A charge of amount 1μC is placed at the centre of square. 

There is repulsion force between charges located at A and O is equal in magnitude but having opposite direction relative to the repulsion force between the charges located at C and O. Hence, they will cancel each other forces. 

Similarly, there is attraction force between charges located at B and O equal in magnitude but having opposite direction relative to the attraction force between the charge placed at D and O. Hence, they also cancel each other forces. 

Therefore, the net force due to the four charges placed at the corners of the square on 1μC charge which is placed at centre O is zero.


6. 

a) Two-point Charges qA=3μC and qB=3μC are Located 20cm Apart in Vacuum. What Is the Electric Field at the Midpoint O of the Line Ab Joining the Two Charges? 

Ans. 

O is the mid-point of line AB. Distance between the two charges i.e., AB=20cm


Distance between the two charges


Therefore, OA=OB=10cm.

Electric field at point O due to +3μC charge:

E1=3×1064πεo(AO)2

E1=3×1064πεo(10×102)2NC1 along OB.

Where, εo is the permittivity of free space.

14πεo=9×109Nm2C2.

Electric field at point O due to 3μC charge:

E2=3×1064πεo(OB)2

E2=3×1064πεo(10×102)2NC1 along OB.

E=E1+E2

E=2×9×109×3×106(10×102)2NC1 

[since, E1 and E2 having same values, so, the value is multiplied with 2]

E=5.4×106NC1 along OB.

Therefore, the electric field at mid-point O is 5.4×106NC1 along OB.


b) If a Negative Test Charge of Magnitude 1.5×109C is Placed at This Point, What is the Force Experienced by the Test Charge?

Ans.

A test charge 1.5×109C is placed at mid-point O.

q=1.5×109C

Force experienced by test charge, F=qE

F=1.5×109×5.4×106

F=8.1×103N

The force is aimed along line OA. The negative test charge is repelled by the charge located at point B but attracted towards A. 

Therefore, the force felt by the test charge is 8.1×103N along OA.


7. A System Has Two Charges qA=2.5×107C and qB=2.5×107C located at points A (0,0,15) and (0,0,15) Respectively. What Are the Total Charge and Electric Dipole Moment of the System?

Ans. 

Two charges are located at their respective position.


Total Charge


The value of charge at A, qA=2.5×107C

The value of charge at B, qB=2.5×107C

Net amount of charge, qnet=qA+qB

qnet=+2.5×1072.5×107

qnet=0

The distance between two charges at A and B,

d=15+15=30cm

d=0.3m

The electric dipole moment of the system is given by

P=qA×d=qB×d

P=2.5×107×0.3

P=7.5×108Cm along z-axis.

Therefore, 7.5×108Cm is the electric dipole moment of the system and it is along positive z-axis.


8. 

a) Two Insulated Charged Copper Spheres a and B Have Their Centres Separated by a Distance of 50cm. What Is the Mutual Force of Electrostatic Repulsion If the Charge on Each is 6.5×107C? The Radii of A and B Are Negligible Compared to the Distance of Separation. 

Ans. 

It is given that:

Charges on both A and B is equal to qA=qB=6.5×107C

Distance between the centres of the spheres is given as r=50cm=0.5m

It is known that the force of repulsion between the two spheres would be

F=qAqB4πε0r2

where,

εo is the permittivity of the free space

Substituting the known values in the above expression, 

F=9×109×(6.5×107)2(0.5)2=1.52×102N

The mutual force of electrostatic repulsion between the two spheres is 1.52×102N.


b)What is the Force of Repulsion if Each Sphere is Charged Double the Above Amount, and the Distance Between Them is Halved?

Ans.

Next, it is told that the charges on both the spheres are doubled and the distance between the centres of the spheres is halved. Thus,

qA=qB=2×6.5×107=13×107C

r=12(0.5)=0.25m

Now, substituting this in the relation for force, 

F=qAqB4πε0r2=9×109×(13×107)2(0.25)2=0.243N

The new mutual force of electrostatic repulsion between the two spheres is 0.243N.


9. Suppose the Spheres A and B in Exercise 1.12 Have Identical Sizes. a Third Sphere of the Same Size but Uncharged Is Brought in Contact With the First, Then Brought in Contact With the Second, and Finally Removed from Both. What Is the New Force of Repulsion Between A and B?

Ans. 

It is given that:

Distance between the spheres A and B is r=0.5m

The charge on each sphere initially is qA=qB=6.5×107C

Now, when uncharged sphere C is made to touch the sphere A, the amount of charge from A will get transferred to the sphere C, making both A and C to have equal charges in them. Clearly,

qA=qC=12(6.5×107)=3.25×107C

Now, when the sphere C is made to touch the sphere B, there is similar transfer of charge making both C and B to have equal charges in them. Clearly,

qC=qB=3.25×107+6.5×1072=4.875×107C

Thus, the new force of repulsion between the spheres A and B will turn out to be

F=qAqB4πε0r2=9×109×3.25×107×4.875×107(0.5)2=5.703×103N


10. Two Large, Thin Metal Plates Are Parallel and Close to Each Other. on Their Inner Faces, the Plates Have Surface Charge Densities of Opposite Signs and of Magnitude 17.0×1022Cm2. What is E in the Outer Region of the First Plate? What is E in the Outer Region of the Second Plate? What is E Between the Plates?

Ans: 

The given nature of metal plates is represented in the figure below: 


metal plates


Here, A and B are two parallel plates kept close to each other. The outer region of plate A is denoted as I, outer region of plate B is denoted as III, and the region between the plates, A and B, is denoted as II.

It is given that:

Charge density of plate A, σ=17.0×1022C/m2 

Charge density of plate B, σ=17.0×1022C/m2

In the regions I and III, electric field E is zero. This is because the charge is not enclosed within the respective plates.

Now, the electric field E in the region II is given by

E=|σ|ε0 

where, 

ε0= Permittivity of free space =8.854×1012N1C2m2 

Clearly,

E=17.0×10228.854×1012

E=1.92×1010N/C

Thus, it can be concluded that the electric field between the plates is 1.92×1010N/C.


11. An Oil Drop of 12 Excess Electrons Is Held Stationary Under a Constant Electric Field of 2.55×104NC1 in Millikan's Oil Drop Experiment. the Density of the Oil Is 1.26gm/cm3. Estimate the Radius of the Drop. (g=9.81ms2,e=1.60×1019C).

Ans: 

It is given that:

The number of excess electrons on the oil drop, n=12 

Electric field intensity, E=2.55×104NC1 

The density of oil, ρ=1.26gm/cm3=1.26×103kg/m3 

Acceleration due to gravity, g=9.81ms2 

Charge on an electron e=1.60×1019C

Radius of the oil drop =r 

Here, the force (F) due to electric field E is equal to the weight of the oil drop (W).

Clearly,

F=W

Eq=mg

Ene=43πr2ρ×g 

where,

q is the net charge on the oil drop =ne

m is the mass of the oil drop =Volume of the oil drop×Density of oil=43πr3×p

Therefore, radius of the oil drop can be calculated as 

r=3Ene4πρg

r=3×2.55×104×12×1.6×10194×3.14×1.26×103×9.81

r=946.09×1021

r=9.72×1010m

Therefore, the radius of the oil drop is 9.72×1010m.


12. In a Certain Region of Space, Electric Field Is Along the Z-Direction Throughout. the Magnitude of Electric Field Is, However, Not Constant but Increases Uniformly Along the Positive Z-Direction, at the Rate of 105NC1 Per Meter. What Are the Force and Torque Experienced by a System Having a Total Dipole Moment Equal to 107Cm in the Negative Z-Direction?

Ans: 

It is known that the dipole moment of the system, P=q×dl=107Cm

Also, the rate of increase of electric field per unit length is given as

 dEdl=105NC1

Now, the force (F) experienced by the system is given by F=qE

F=qdEdl×dl

F=PdEdl

F=107×105

F=102N 

Clearly, the force is equal to 102N in the negative z-direction i.e., it is opposite to the direction of electric field. 

Thus, the angle between electric field and dipole moment is equal to 180.

Now, the torque is given by τ=PEsinθ

𝜏 = PEsin1800

Therefore, it can be concluded that the torque experienced by the system is zero.


13. 

a) A Conductor With a Cavity as Shown in the Fig. 1.36(a) Is Given a Charge Q. Show That the Entire Charge Must Appear on the Outer Surface of the Conductor. 


Conductor


Ans: 

Firstly, let us consider a Gaussian surface that is lying within a conductor as a whole and enclosing the cavity. Clearly, the electric field intensity E inside the charged conductor is zero.

Now, let q be the charge inside the conductor and ε0, the permittivity of free space.

According to Gauss's law,

Flux is given by

 ϕ=E.ds=qε0 

Here, ϕ=0  as E=0 inside the conductor

Clearly, 

0=q8.854×1012

q=0 

Therefore, the charge inside the conductor is zero.

And hence, the entire charge Q appears on the outer surface of the conductor.


b) Another Conductor B With Charge q is Inserted Into Cavity Keeping B Insulated from A. Show That the Total Charge on the Outside Surface of A is Q+q [Fig. 1.36 (b)].

Ans.

The outer surface of conductor A has a charge of Q.

It is given that another conductor B, having a charge +q is kept inside conductor A and is insulated from the conductor A.

Clearly, a charge of q will get induced in the inner surface of conductor A and a charge of +q will get induced on the outer surface of conductor A.

Therefore, the total charge on the outer surface of conductor A amounts to Q+q.


C) a Sensitive Instrument is to Be Shielded from the Strong Electrostatic Fields in Its Environment. Suggest a Possible Way.

Ans.

A sensitive instrument can be shielded from a strong electrostatic field in its environment by enclosing it fully inside a metallic envelope. 

Such a closed metallic body provides hindrance to electrostatic fields and thus can be used as a shield.


14. A Hollow Charged Conductor Has a Tiny Hole Cut Into Its Surface. Show That the Electric Field in the Hole Is [σ2ε0]n, where n is the Unit Vector in the Outward Normal Direction, and σ Is the Surface Charge Density Near the Hole.

Ans: 

Firstly, let us consider a conductor with a cavity or a hole. It is known that the electric field inside the cavity is zero. 

Let us assume E to be the electric field just outside the conductor, q be the electric charge, σ be the charge density, and ε0, the permittivity of free space.

We know that charge |q|=σ×d 

Now, according to Gauss's law,

ϕ=E.ds=|q|ε0

E.ds=σ×dε0

E=σε0n

where n is the unit vector in the outward normal direction.  

Thus, the electric field just outside the conductor is σε0n. Now, this field is actually a superposition of the field due to the cavity E1 and the field due to the rest of the charged conductor E2. These electric fields are equal and opposite inside the conductor whereas equal in magnitude as well as direction outside the conductor. Clearly,

E1+E2=E

E1=E2=E2=σ2ε0n

Therefore, the electric field in the hole is  σ2ε0n

Hence, proved.


15. Obtain the Formula for the Electric Field Due to a Long Thin Wire of Uniform Linear Charge Density λ Without Using Gauss's Law. (hint: Use Coulomb's Law Directly and Evaluate the Necessary Integral)

Ans: 

Firstly, let us take a long thin wire XY as shown in the figure below. This wire is of uniform linear charge density λ.


long thin wire


Now, consider a point A at a perpendicular distance l from the mid-point O of the wire as shown in the figure below:


triangular graph


Consider E to be the electric field at point A due to the wire.

Also consider a small length element dx on the wire section with OZ=x as shown.

Let q be the charge on this element.

Clearly, q=λdx 

Now, the electric field due to this small element can be given as

dE=14πε0λdx(AZ)2 

However, AZ=12+x2 

dE=λdx4πε0(12+x2)

Now, let us resolve the electric field into two rectangular components. Doing so, dEcosθ is the perpendicular component and dEsinθ is the parallel component.

When the whole wire is considered, the component dEsinθ gets cancelled and only the perpendicular component dEcosθ  affects the point A.

Thus, the effective electric field at point A due to the element dx can be written as 

dE1=λdxcosθ4πε0(l2+x2)    ....(1) 

Now, in ΔAZO, we have

tanθ=xl

x=ltanθ ......(2) 

On differentiating equation (2), we obtain

dx=lsec2dθ ......(3) 

From equation (2)

x2+l2=l2+l2tan2θ

l2(1+tan2θ)=l2sec2θ

x2+l2=l2sec2θ .....(4)

Putting equations (3) and (4) in equation (1), we obtain

dE1=λlsec2dθ4πε0(l2sec2θ)cosθ

dE1=λcosθdθ4πε0l .....(5)

Now, the wire is taken so long that ends from π2 to +π2.

Therefore, by integrating equation (5), we obtain the value of field E1 as

π2π2dE1=π2π2λ4πε0lcosθdθ

E1=λ4πε0l×2

E1=λ2πε0l

Thus, the electric field due to the long wire is derived to be equal to λ2πε0l.


16. it Is Now Believed That Protons and Neutrons (which Constitute Nuclei of Ordinary Matter) Are Themselves Built Out of More Elementary Units Called Quarks. a Proton and a Neutron Consist of Three Quarks Each. Two Types of Quarks, the So Called 'up Quark (denoted by u)  of Charge (+12)e and the 'down' Quark (denoted by d) of Charge (13)e Together With Electrons Build up Ordinary Matter. (Quarks of Other Types Have Also Been Found Which Give Rise to Different Unusual Varieties of Matter.) Suggest a Possible Quark Composition of a Proton and Neutron.                       

Ans: 

It is known that a proton has three quarks. Let us consider n up quarks in a proton, each having a charge of +(23e)

Now, the charge due to n up quarks =(23e)n 

The number of down quarks in a proton =3n 

Also, each down quark has a charge of 13e 

Therefore, the charge due to (3n) down quarks =(13e)(3n)

We know that the total charge on a proton =+e 

Therefore,

e=(23e)n+(13e)(3n)

e=(2ne3)e+ne3

2e=ne

n=2

Clearly, the number of up quarks in a proton, n=2

Thus, the number of down quarks in a proton =3n=32=1 

Therefore, a proton can be represented as uud.

A neutron is also said to have three quarks. Let us consider n up quarks in a neutron, each having a charge of +(23e) .

It is given that the charge on a neutron due to n up quarks =(+32e)n 

Also, the number of down quarks is (3n), each having a charge of =(32)e

Thus, the charge on a neutron due to (3n) down quarks =(13e)(3n)

Now, we know that the total charge on a neutron =0 

Thus,

0=(23e)n+(13e)(3n)

0=(2ne3)e+ne3

e=ne

n=1

Clearly, the number of up quarks in a neutron, n=1

Thus, the number of down quarks in a neutron =3n=2 

Therefore, a neutron can be represented as udd.


17. 

a) Consider an Arbitrary Electrostatic Field Configuration. A Small Test Charge Is Placed at a Null Point (i.e., where E=0) Of the Configuration. Show That the Equilibrium of the Test Charge Is Necessarily Unstable.

Ans:

Firstly, let us assume that the small test charge placed at the null point of the given setup is in stable equilibrium. 

By stable equilibrium, it means that even a slight displacement of the test charge in any direction will cause the charge to return to the null point as there will be strong restoring forces acting around it. 

This further suggests that all the electric lines of force around the null point act inwards and towards the given null point. 

But by Gauss law, we know that the net electric flux through a chargeless enclosing surface is equal to zero. This truth contradicts the assumption which we had started with. Therefore, it can be concluded that the equilibrium of the test charge is necessarily unstable.


b) Verify this Result for the Simple Configuration of Two Charges of the Same Magnitude and Sign Placed at a Certain Distance Apart.

Ans.

When we consider this configuration setup with two charges of the same magnitude and sign placed at a certain distance apart, the null point happens to be at the mid-point of the line joining these two charges. 

As per the previous assumption, the test charge, when placed at this mid-point will experience strong restoring forces when it tries to displace itself. 

But when the test charge tries to displace in a direction normal to the line joining the two charges, the test charge gets pulled off as there is no restoring force along the normal to the line considered. 

Since stable equilibrium prioritizes restoring force in all directions, the assumption in this case also gets contradicted. 


18. A Particle of Mass m and Charge (q) Enters the Region Between the Two Charged Plates Initially Moving Along X- Axis With Speed vx (like particle 1 in Fig 1.33). The Length of Plate is L and a Uniform Electric Field E Is Maintained Between the Plates. Show That the Vertical Deflection of the Particle at the Far Edge of the Plate is qEL22mvx2.

Compare This Motion with Motion of a Projectile in Gravitational Field Discussed in Section 4.10 of Class Xi Textbook of Physics.

Ans: 

It is given that:

The charge on a particle of mass m=q

Velocity of the particle =vx 

Length of the plates =L 

Magnitude of the uniform electric field between the plates =E 

Mechanical force, F= Mass (m)×Acceleration (a) 

Thus, acceleration, a=Fm

However, electric force, F=qE 

Therefore, acceleration, =qEm  .........(1)

Here, the time taken by the particle to cross the field of length L is given by,

t=Length of the plateVelocity of the plate=Lvx  ......(2)

In the vertical direction, we know that the initial velocity, u=0 

Now, according to the third equation of motion, vertical deflection s of the particle can be derived as

s=ut+12at2

s=0+12(qEm)(Lvx)2 

s=qEL22mvx2   .....(3)

Therefore, the vertical deflection of the particle at the far edge of the plate is qEL22mvx2 


Important Formulas from Class 12 Physics Chapter 1 Electric Charges And Fields

  • Coulomb's Law: F=kq1q2r2

  • Electric Field: E=Fq

  • Electric Dipole Moment: p=qd

  • Potential Energy of a Dipole: U=pEcosθ

  • Electric Flux: ϕ=EAcosθ

  • Gauss's Law: EdA=qenclosedϵ0


Benefits of Learning with Class 12 Physics  Chapter 1 Electric Charges And Fields

  • Builds a strong conceptual foundation for advanced topics in electromagnetism.

  • Enhances problem-solving skills required for board exams and competitive tests like JEE and NEET.

  • Develops analytical skills through derivations and applications of formulas.

  • Provides insight into real-world phenomena such as electric fields and charge distributions.

  • Helps in understanding practical applications like capacitors and charge conservation.

  • Strengthens mathematical skills by solving complex numerical problems.

  • Prepares students for higher studies in physics and engineering fields.


Tips to Study Class 12 Chapter 1 Electric Charges and Fields Important Questions 

  1. Start by thoroughly reading the NCERT textbook to build a solid conceptual base.

  2. Memorise key formulas such as Coulomb's law and Gauss's law, and practice their derivations.

  3. Solve all NCERT examples and exercises to strengthen your understanding.

  4. Focus on understanding the applications of electric field lines and flux.

  5. Regularly revise important concepts like electric dipole and potential energy.

  6. Solve previous years' board and competitive exam questions related to this chapter.

  7. Practice numerical problems daily to improve accuracy and speed.

  8. Create concise notes summarising formulas, key points, and derivations for quick revision.


Related Study Materials for CBSE Class 12 Physics  Chapter 1


Conclusion

"Electric Charges and Fields" is a pivotal chapter in Class 12 Physics, bridging the gap between theoretical concepts and practical applications. Students can master this chapter effectively by leveraging NCERT solutions, practicing important questions, and using the FREE PDF provided by Vedantu. Stay consistent, focus on understanding, and success will follow.


CBSE Class 12 Physics Chapter-wise Important Questions

CBSE Class 12 Physics Chapter-wise Important Questions and Answers cover topics from all other chapters, helping students prepare thoroughly by focusing on key topics for easier revision.



Additional Study Materials for Class 12 Physics 

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FAQs on Electric Charges and Fields Class 12 Important Questions: CBSE Physics Chapter 1

1. Where can I find Electric Charges and Fields Class 12 Important Questions?

Vedantu provides selected questions for each chapter of the NCERT textbook which are important for exam preparation. Students can avail the important questions PDF file for Class 12 Physics Chapter 1 Electric Charges and Fields on Vedantu’s platform. The free PDF version allows students to download these files anytime, anywhere. The list of important questions is picked with great efforts by expert Physics tutors to help students in scoring well in the paper. They also offer the solutions as per the board guidelines and exam pattern. Students must practice these questions in order to revise all types of questions and get an idea of what might be asked in exams.

2. Why should I practice Vedantu’s important questions for Class 12 Physics Chapter 1?

Many students find it challenging to score well in Class 12 Physics paper. However, with a thorough understanding of concepts and regular practice, scoring in the Physics paper won’t be difficult. For the better practice of Class 12 Physics Chapter 1 as well as other chapters, students can download the free PDF file of important questions from Vedantu’s site. These questions are added as per the exam pattern and the latest syllabus. This will ensure that you have better coverage on the concepts as well as a fair idea of what types of questions to expect in the exam. You can also find solutions for these questions provided by experts in case they are having trouble solving them.

3. What are some of the important questions for Class 12 Physics Chapter 1?

Following are some of the important questions for Class 12 Physics Chapter 1 - Electric Charges and Fields:

  • A charged rod P attracts rod R whereas P repels another charged rod Q. What type of force is developed between Q and R?

  • Define one coulomb.

  • No two electric lines of force can intersect each other? Why?

  • Explain why two field lines never cross each other at any point.

  • What is the force between two small charged spheres having charges of 2 x 10-7 C and 3 x 10-3 C placed 30 cm apart in the air?

For solutions to these questions as well as more important questions on Class 12 Physics Chapter 1, students can refer to Vedantu’s important questions for Class 12 Physics Chapter 1.

4. What are some of the important contents of Chapter 1 Electric Charges and Fields of Class 12 Physics?

Class 12 Physics Chapter 1 Electric Charges and Fields consists of questions on the linear charge density of an infinite line charge and many other questions regarding the electric field. In this chapter, students will be solving questions on how to calculate forces between two charged particles which are kept at a certain distance from each other. Students will be learning about the theory behind the charge appearing in materials due to rubbing against each other just like when we rub a glass rod with a silk cloth or a dry comb on our hair. There are many other interesting concepts in this chapter.

5. How are Electric Charges and Fields Class 12 Important Questions beneficial for students?

Important questions of Class 12 Physics Chapter 1 available at Vedantu app and website can be extremely useful for the students of Class 12. They can use the important questions for their board exam preparation and can also prepare for their entrance exam. The important questions can help them to understand the concepts easily because all important questions are prepared using basic and simple language that makes it easy for the students to understand the facts. 

6. How can I download important questions for Chapter 1- Electric Charges and Fields of Class 12 Physics from Vedantu?

Students can download the Chapter 1- Electric Charges and Field of Class 12 Physics important questions from Vedantu free of cost. They have to visit the website and search for important questions of Chapter 1 of Class 12 Physics then click on the download link. They can save the PDF file on their computers and can use the important questions to prepare for their final exams and entrance exam. All questions are important for the board exams and NEET and JEE Exams.

7. What are the main concepts students will study in Chapter 1 of Class 12 Physics?

Students will study different concepts related to Electric Charges and Fields in Class 12 Physics Chapter 1. They will study Electrostatic capacitance and potential, current electricity, the relation between moving charges and magnetism, electromagnetic induction, alternating current, and electromagnetic waves. Students must understand all the concepts given and can refer to the important questions given on Vedantu for a clear understanding of the concepts.

8. What is an electric charge according to Chapter 1 of Class 12 Physics?

Electric charge is a physical property of matter. The matter experiences a force when it is placed in an electromagnetic field. Electric charges can be positive or negative. The particle that carries a negative charge is called an electron and the particle that carries a positive charge is called a proton. Many experiments were done for analyzing electric charge. The experiment revealed that like charges move away from each other and unlike charges come closer to each other. 

9. Write a short note on the electric field related to Chapter 1 of Class 12 Physics?

The electric field is a field that surrounds an electric charge. It exerts a force on other charges present in the field. The direction of the force exerted on the charge is the same as the direction of the field. An electric field is believed to be a result of electromagnetic force. If the direction of the force is outward it is considered positive and if the direction of the force is inward it is considered negative. 

10. What are the important topics in electric charges and fields 12?

Important topics in Electric Charges and Fields Class 12 include electric charge, Coulomb's law, electric field and electric field lines, Gauss's law, electric potential, capacitance, and conductors and insulators.

11. What is charge class 12 questions?

Questions on charge in Class 12 typically cover topics such as the properties of charge, conservation of charge, Coulomb's law, and the behavior of charged particles in electric fields.

12. Is the electric field due to a charge configuration with total charge zero necessarily?

No, the electric field due to a charge configuration with total charge zero is not necessarily zero. The electric field depends on the distribution of charges, and even if the total charge is zero, the electric field may not be zero due to the arrangement of charges.