Question

The electric field is measured at point P (0,0,d) generated due to various charge distributions and dependence of E on d is found to be different for different charges. List 1 contains different relations between E and d. list 2 describes different electric charge distributions, along with their locations. Match the functions in list 1 with related charge distribution in list 2.

List 1	List 2
(P) E is independent of d	(1) A point charge on the origin
(Q) \[E\propto \dfrac{1}{d}\]	(2) A small dipole with point charges Q (0,0,l) and -Q at (0,0,-l). Take \[2l\ll d\]
(R) \[E\propto \dfrac{1}{{{d}^{2}}}\]	(3) an infinite line of charge coincident with the X-axis, with uniform linear charge density \[\lambda \]
(S) \[E\propto \dfrac{1}{{{d}^{3}}}\]	(4) Two infinite wires carrying uniform linear charge density parallel to the X-axis. The one along (y=0,z=l) has a charge density + \[\lambda \] and one along (y=0,z=-l) has charge density - \[\lambda \]. Take \[2l\ll d\]
	(5) Infinite plane charge coincident with the xy plane with uniform surface charge density

A. \[P\to 5;Q\to 3;R\to 1;S\to 2\]
B. \[P\to 5;Q\to 3;R\to 1,4;S\to 2\]
C. \[P\to 5;Q\to 3;R\to 1,2;S\to 4\]
D. \[P\to ;Q\to 2;R\to 1;S\to 5\]

Answer 1

Hint: In this question, we have been asked to match the options in list 1 with appropriate answers from list 2. In list 1 we have been given the different relations between electric field and distance d and list 2 describes different electric charge distributions, along with their locations. Therefore, to solve this question we shall calculate the electric field for electric charge distribution given in list 2. We shall then match our answers with relation given in list 1.

Complete answer:
Considering the first point given in list 2
Therefore, we know the electric field for point charge is given by,
\[E=k\dfrac{{{q}^{2}}}{{{d}^{2}}}\]
From the above equation we can say that,
\[E\propto \dfrac{1}{{{d}^{2}}}\]
Therefore,
\[R\to 1\] ………………. (1)
Now, in second row of list 2 it is given that the charge distribution is of a small dipole
Therefore, electric field for a small dipole with point charge is given by,
\[E=k\dfrac{\rho }{{{d}^{3}}}\]
Where,
\[\rho \] is the dipole moment \[\rho =q\times 2l\]. It is given that \[2l\ll d\].
Therefore, we can say that,
\[E\propto \dfrac{1}{{{d}^{3}}}\]
Therefore,
\[S\to 2\] …………….. (2)
Now, the third row from list 2 gives a distribution of infinite line charge with uniform charge density,
Therefore, the electric field for infinite line charge is given by,
\[E=2k\dfrac{\lambda }{d}\]
Where,
\[\lambda \] is the uniform charge density
Therefore, we can say that,
\[E\propto \dfrac{1}{d}\]
Therefore,
\[Q\to 3\] ……………….. (3)
The fourth statement in list 2 has a distribution of two infinite wire carrying charges with uniform density.
We know that electric field due to infinite wire is given by,
\[E=2k\dfrac{\lambda }{d+l}\]
Therefore, electric field for first wire that is along (y=0,z=l) is given by,
\[E=2k\dfrac{\lambda }{d+l}\] ……………. (A)
Similarly, for second wire which is along (y=0,z=-l)
\[E=2k\dfrac{\lambda }{d-l}\] …………….. (B)
Now, from (A) and (B)
We get,
\[E=2k\lambda \left( \dfrac{1}{d-l}-\dfrac{1}{d+l} \right)\]
On solving we get,
\[E=\dfrac{2k\lambda \times 2l}{{{d}^{2}}\left( 1-\dfrac{{{l}^{2}}}{{{d}^{2}}} \right)}\]
Therefore, from above equation we can say that,
\[E\propto \dfrac{1}{{{d}^{2}}}\]
Therefore,
\[R\to 4\] ………………… (4)
Now, the last statement in list 2 has a charge distribution along an infinite plane or sheet,
We know electric field for infinite sheet is given by,
\[E=\dfrac{\sigma }{2{{\varepsilon }_{0}}}\]
Therefore, we can say that electric field for infinite sheet with uniform charge density is independent of distance d
Therefore,
\[P\to 5\]………………. (5)
Therefore, from (1), (2), (3), (4) and (5) we can say that,

The correct answer is option B.

Note:
The ratio of electric force per unit charge is defined as the electric field. The direction of the force it will exert on a positive charge is taken as the direction of the electric field. The electric field from a positive charge is always radially outwards and vice versa. Electric field is a physical field that surrounds the electric charge and exerts the force on the other charges within the field.