
A point charge of charge q and mass m is placed at rest at point X at distance r from a short electric dipole. The initial acceleration of charge ${[ a=\dfrac{k\times q\times p}{2mr^3}n ]}$, where ${[ k=\dfrac{1}{4\pi\varepsilon_0} ]}$ then n is-
A. 6.4
B. 2.6
C. 5.0
D. 8.9
Answer
162.3k+ views
Hint:As we know the force is mass multiplied by acceleration but in the gravitational force the acceleration is acceleration due to gravity so acceleration does not depend on the mass and in electric force the acceleration depends on the mass. As we break the component P it breaks into component Ep1 and Ep2 axial and equatorial respectively.
Formula used:
${f=m\times a}$
${f=q\times\vec{E}}$
${\left|\vec{E}\right|=\sqrt{\left|\vec{E}p_1\right|^2+\left|\vec{Ep_2}\right|^2}}$
Complete answer:
As we know force is-
${f=m\times a}$
and, ${f=q\times\vec{E}}$
where, m = mass
a = acceleration
q = given charge
E = electric field
Now to find out the net electric field from the dipole first we need to divide p in two parts which is p1 and p2 whose values are cosϴ and sinϴ respectively.
p1 = cos ϴ
p2 = sin ϴ
These are the two components of our dipole equatorial and axial, equatorial position will be at the 180˚ of the p2 which is Ep2 and along the axis of p1 will be Ep1.
Now for equatorial position our electric field will be-
${{\vec{E}}_{(equatorial)}=\dfrac{pk}{r^3}}$
${Since,[ k=\dfrac{1}{4\pi\varepsilon_0} ]so,}$
${{\vec{E}}_{(equatorial)}=\dfrac{p}{r^3}\times\dfrac{1}{4\pi\varepsilon_0}}$
${{\vec{E}}_{p_2}=\dfrac{p}{r^3}\times\dfrac{1}{4\pi\varepsilon_0}}$
${{\vec{E}}_{(equatorial)}=\dfrac{p\ sin\theta}{r^3}\times\dfrac{1}{4\pi\varepsilon_0}}$
As we know that the angle ϴ is 60˚ so
${Sin\ 60˚ =32 }$
${{\vec{E}}_{(equatorial)}=\dfrac{p\ }{r^3}\times\dfrac{1}{4\pi\varepsilon_0}\times\dfrac{\sqrt3}{2}}$
Now for axial position our electric field will be-
${{\vec{E}}_{(axial)}=\dfrac{2pk}{r^3}}$
${Since, [ k=\dfrac{1}{4\pi\varepsilon_0} ] so,}$
${{\vec{E}}_{(equatorial)}=\dfrac{2p}{r^3}\times\dfrac{1}{4\pi\varepsilon_0}}$
${{\vec{E}}_{p_2}=\dfrac{2p}{r^3}\times\dfrac{1}{4\pi\varepsilon_0}}$
${{\vec{E}}_{(equatorial)}=\dfrac{2p\ cos\theta}{r^3}\times\dfrac{1}{4\pi\varepsilon_0}}$
As we know that the angle ϴ is 60˚ so
${cos\ 60˚ =12 }$
${{\vec{E}}_{(equatorial)}=\dfrac{2p\ }{r^3}\times\dfrac{1}{4\pi\varepsilon_0}\times\dfrac{1}{2}}$
${{\vec{E}}_{(equatorial)}=\dfrac{p\ }{r^3}\times\dfrac{1}{4\pi\varepsilon_0}}$
To determine magnitude of acceleration first we will determine with the help of magnitude of ${force-\left|\vec{f}\right|=m\times\left|\vec{a}\right| }$ -------(i)
${\left|\vec{f}\right|=q\times\left|\vec{E}\right|}$ -------(ii)
Equating both the equations of force (i) and (ii)
${m\times\left|\vec{a}\right|=q\times\left|\vec{E}\right| }$
${\left|\vec{a}\right|=\dfrac{q\times\left|\vec{E}\right|}{m} }$ ---------(iii)
Now magnitude of E will be,
${\left|\vec{E}\right|=\sqrt{\left|\vec{E}p_1\right|^2+\left|\vec{Ep_2}\right|^2}}$
${\left|\vec{E}\right|=\sqrt{\dfrac{p^2k^2}{r^6}+\dfrac{3}{4}\dfrac{p^2k^2}{r^6}}}$
${\left|\vec{E}\right|=\dfrac{pk}{r^3}\sqrt{1+\dfrac{3}{4}}}$
${\left|\vec{E}\right|=\dfrac{pk}{r^3}\sqrt{\dfrac{7}{4}}}$
${\left|\vec{E}\right|=\dfrac{pk}{2r^3}\sqrt7}$
Now, from equation (iii) putting value of mode E acceleration will be-
${\left|\vec{a}\right|=\dfrac{q\times p\times k}{2mr^3}\sqrt7 }$ --------(iv)
Now as given in question
${a=\dfrac{k\times q\times p}{2mr^3}n }$ -------(v)
Now equating both the equation (v) and(iv)
${\dfrac{k\times q\times p}{2mr^3}n=\dfrac{q\times p\times k}{2mr^3}\sqrt7}$
${n=\sqrt7}$
${n=2.645}$
Thus, the value of n is 2.6 and correct option is (B)
Note:Axial line is the line that passes through both the poles of the dipole the positive and the negative poles so basically it is through its axis, so it is called axial component and equatorial component is the component that is perpendicular to axial component.
Formula used:
${f=m\times a}$
${f=q\times\vec{E}}$
${\left|\vec{E}\right|=\sqrt{\left|\vec{E}p_1\right|^2+\left|\vec{Ep_2}\right|^2}}$
Complete answer:

As we know force is-
${f=m\times a}$
and, ${f=q\times\vec{E}}$
where, m = mass
a = acceleration
q = given charge
E = electric field
Now to find out the net electric field from the dipole first we need to divide p in two parts which is p1 and p2 whose values are cosϴ and sinϴ respectively.
p1 = cos ϴ
p2 = sin ϴ
These are the two components of our dipole equatorial and axial, equatorial position will be at the 180˚ of the p2 which is Ep2 and along the axis of p1 will be Ep1.
Now for equatorial position our electric field will be-

${{\vec{E}}_{(equatorial)}=\dfrac{pk}{r^3}}$
${Since,[ k=\dfrac{1}{4\pi\varepsilon_0} ]so,}$
${{\vec{E}}_{(equatorial)}=\dfrac{p}{r^3}\times\dfrac{1}{4\pi\varepsilon_0}}$
${{\vec{E}}_{p_2}=\dfrac{p}{r^3}\times\dfrac{1}{4\pi\varepsilon_0}}$
${{\vec{E}}_{(equatorial)}=\dfrac{p\ sin\theta}{r^3}\times\dfrac{1}{4\pi\varepsilon_0}}$
As we know that the angle ϴ is 60˚ so
${Sin\ 60˚ =32 }$
${{\vec{E}}_{(equatorial)}=\dfrac{p\ }{r^3}\times\dfrac{1}{4\pi\varepsilon_0}\times\dfrac{\sqrt3}{2}}$
Now for axial position our electric field will be-
${{\vec{E}}_{(axial)}=\dfrac{2pk}{r^3}}$
${Since, [ k=\dfrac{1}{4\pi\varepsilon_0} ] so,}$
${{\vec{E}}_{(equatorial)}=\dfrac{2p}{r^3}\times\dfrac{1}{4\pi\varepsilon_0}}$
${{\vec{E}}_{p_2}=\dfrac{2p}{r^3}\times\dfrac{1}{4\pi\varepsilon_0}}$
${{\vec{E}}_{(equatorial)}=\dfrac{2p\ cos\theta}{r^3}\times\dfrac{1}{4\pi\varepsilon_0}}$
As we know that the angle ϴ is 60˚ so
${cos\ 60˚ =12 }$
${{\vec{E}}_{(equatorial)}=\dfrac{2p\ }{r^3}\times\dfrac{1}{4\pi\varepsilon_0}\times\dfrac{1}{2}}$
${{\vec{E}}_{(equatorial)}=\dfrac{p\ }{r^3}\times\dfrac{1}{4\pi\varepsilon_0}}$
To determine magnitude of acceleration first we will determine with the help of magnitude of ${force-\left|\vec{f}\right|=m\times\left|\vec{a}\right| }$ -------(i)
${\left|\vec{f}\right|=q\times\left|\vec{E}\right|}$ -------(ii)
Equating both the equations of force (i) and (ii)
${m\times\left|\vec{a}\right|=q\times\left|\vec{E}\right| }$
${\left|\vec{a}\right|=\dfrac{q\times\left|\vec{E}\right|}{m} }$ ---------(iii)
Now magnitude of E will be,
${\left|\vec{E}\right|=\sqrt{\left|\vec{E}p_1\right|^2+\left|\vec{Ep_2}\right|^2}}$
${\left|\vec{E}\right|=\sqrt{\dfrac{p^2k^2}{r^6}+\dfrac{3}{4}\dfrac{p^2k^2}{r^6}}}$
${\left|\vec{E}\right|=\dfrac{pk}{r^3}\sqrt{1+\dfrac{3}{4}}}$
${\left|\vec{E}\right|=\dfrac{pk}{r^3}\sqrt{\dfrac{7}{4}}}$
${\left|\vec{E}\right|=\dfrac{pk}{2r^3}\sqrt7}$
Now, from equation (iii) putting value of mode E acceleration will be-
${\left|\vec{a}\right|=\dfrac{q\times p\times k}{2mr^3}\sqrt7 }$ --------(iv)
Now as given in question
${a=\dfrac{k\times q\times p}{2mr^3}n }$ -------(v)
Now equating both the equation (v) and(iv)
${\dfrac{k\times q\times p}{2mr^3}n=\dfrac{q\times p\times k}{2mr^3}\sqrt7}$
${n=\sqrt7}$
${n=2.645}$
Thus, the value of n is 2.6 and correct option is (B)
Note:Axial line is the line that passes through both the poles of the dipole the positive and the negative poles so basically it is through its axis, so it is called axial component and equatorial component is the component that is perpendicular to axial component.
Recently Updated Pages
Fluid Pressure - Important Concepts and Tips for JEE

JEE Main 2023 (February 1st Shift 2) Physics Question Paper with Answer Key

Impulse Momentum Theorem Important Concepts and Tips for JEE

Graphical Methods of Vector Addition - Important Concepts for JEE

JEE Main 2022 (July 29th Shift 1) Chemistry Question Paper with Answer Key

JEE Main 2023 (February 1st Shift 1) Physics Question Paper with Answer Key

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Charging and Discharging of Capacitor

Wheatstone Bridge for JEE Main Physics 2025

Formula for number of images formed by two plane mirrors class 12 physics JEE_Main

In which of the following forms the energy is stored class 12 physics JEE_Main
