Question

Four points \[a,b,c \] and \[d \] are set at equal distance from the center of a dipole as shown in the figure. The electrostatic potential \[{V_a},{V_b},{V_c} \] and ​ \[{V_d} \] would satisfy the following relation:

A. \[{V_a} > {V_b} > {V_c} > {V_d} \]
B. \[{V_a} > {V_b} = {V_d} > {V_c} \]
C. \[{V_a} = {V_c} > {V_b} = {V_d} \]
D. \[{V_b} = {V_d} > {V_a} > {V_c} \]

Answer 1

Hint:It is given that four points \[a,b,c \] and \[d \] are set at equal distance from the center of a dipole. Then the expression for electrostatic potential of a charge placed at a point distant \[r \] from the origin is given by \[V = \dfrac{1}{{4\pi {\varepsilon _0}k}}\dfrac{q}{r} \] where, \[{\varepsilon _0} \] is the permittivity of vacuum, \[r \] is the distance of the charge and \[q \] is the charge (either positive or negative).
Electrostatic potential due to positive source charge is positive and negative source charge is negative.

Formula Used:
Electrostatic potential of a charge placed at a point distant \[r \] from the origin is given by
 \[V = \dfrac{1}{{4\pi {\varepsilon _0}k}}\dfrac{q}{r} \]

Complete step by step solution:
 \[a,b,c \] and \[d \] are the four points. Let the electrostatic potential generated by them be and \[{V_d} \] respectively. Let the distance of positive charge \[ + q \] and negative charge \[ - q \] be from the center of the dipole be \[l \] . Let the four points \[a,b,c \] and \[d \] be set at equal distance \[r \] from the center of a dipole.

So, the distance of charge point \[a \] from charge \[ + q \] is \[ = r - l \] , and that from charge \[ - q \] is \[ = r
+ l \] . The value of electrostatic potential \[{V_a} \] due to charge \[ + q \] will be
 \[{V_a} = \dfrac{1}{{4\pi {\varepsilon _0}k}}\dfrac{{ + q}}{{(r - l)}} \] or \[{V_a} = \dfrac{q}{{4\pi {\varepsilon _0}k}}\dfrac{1}{{(r - l)}} \] \[ \to (1) \]
And that due to charge \[ - q \] will be
 \[{V_a} = \dfrac{1}{{4\pi {\varepsilon _0}k}}\dfrac{{ - q}}{{(r + l)}} \] or \[{V_a} = \dfrac{q}{{4\pi {\varepsilon _0}k}}\dfrac{{ - 1}}{{(r + l)}} \] \[ \to (2) \]
The total potential at point \[a \] will be (Adding equations (1) and (2) )
$\Rightarrow {V_a} = \dfrac{q}{{4\pi {\varepsilon _0}k}}\left[ {\dfrac{1}{{(r - l)}} - \dfrac{1}{{(r + l)}}} \right] \\
\Rightarrow \dfrac{1}{{4\pi {\varepsilon _0}k}}\left[ {\dfrac{{(r + l) - (r - l)}}{{(r - l)(r + l)}}} \right] \\
\therefore {V_a} = \dfrac{1}{{4\pi {\varepsilon _0}k}}\left[ {\dfrac{{2l}}{{{r^2} - {l^2}}}} \right] \to (3)$

Similarly, the distance of charge point \[c \] from charge \[ + q \] is \[ = r + l \] , and that from charge \[ - q \] is \[ = r - l \] .

The value of electrostatic potential \[{V_c} \] due to charge \[ + q \] will be
 \[{V_c} = \dfrac{1}{{4\pi {\varepsilon _0}k}}\dfrac{{ + q}}{{(r + l)}} \] or \[{V_c} = \dfrac{q}{{4\pi
{\varepsilon _0}k}}\dfrac{1}{{(r + l)}} \] \[ \to (4) \]
And that due to charge \[ - q \] will be
 \[{V_c} = \dfrac{1}{{4\pi {\varepsilon _0}k}}\dfrac{{ - q}}{{(r - l)}} \] or \[{V_c} = \dfrac{q}{{4\pi
{\varepsilon _0}k}}\dfrac{{ - 1}}{{(r - l)}} \] \[ \to (5) \]
The total potential at point \[a \] will be (Adding equations (4) and (5) )
$\Rightarrow {V_c} = \dfrac{q}{{4\pi {\varepsilon _0}k}}\left[ {\dfrac{1}{{(r + l)}} - \dfrac{1}{{(r - l)}}} \right]$
$\Rightarrow \dfrac{1}{{4\pi {\varepsilon _0}k}}\left[ {\dfrac{{(r - l) - (r + l)}}{{(r + l)(r - l)}}} \right]$ $\therefore {V_c} = \dfrac{1}{{4\pi {\varepsilon _0}k}}\left[ {\dfrac{{ - 2l}}{{{r^2} - {l^2}}}} \right] \to (6) \\$
According to equation (3) and equation (6), the magnitude of electrostatic potential is the same for points \[a \] and \[c \] . Therefore, \[{V_a} = {V_c} \] .
Points \[b \] and \[d \] are equidistant from both the charges, positive as well as negative. Therefore, the electrostatic potential \[{V_b} \] and \[{V_d} \] will be zero. Thus, \[{V_b} = {V_d} \] . Since, \[{V_a} \] and \[{V_c} \] are greater than zero, therefore \[{V_a} = {V_c} > {V_b} = {V_d} \]

Hence, option (C) is the correct answer.

Note: The electrostatic potential at \[{V_b} \] and \[{V_d} \] is taken as zero since points \[b \] and \[d \] are equidistant from both the positive and negative charges. This can also be worked by applying the formula.

Let the distance of points \[b \] and \[d \] from the charges be \[r \] (Refer the diagram below). Then, the electrostatic potential \[{V_b} \] will be
 \[{V_b} = \dfrac{1}{{4\pi {\varepsilon _0}k}}\dfrac{{ + q}}{r} \] or \[{V_b} = \dfrac{q}{{4\pi {\varepsilon
_0}k}}\dfrac{1}{r} \] \[ \to (7) \]
And that due to charge \[ - q \] will be
 \[{V_b} = \dfrac{1}{{4\pi {\varepsilon _0}k}}\dfrac{{ - q}}{r} \] or \[{V_a} = \dfrac{q}{{4\pi {\varepsilon
_0}k}}\dfrac{{ - 1}}{r} \] \[ \to (8) \]
The total potential at point \[a \] will be (Adding equations (7) and (8) )
$\Rightarrow {V_b} = \dfrac{q}{{4\pi {\varepsilon _0}k}}\left[ {\dfrac{1}{r} - \dfrac{1}{r}} \right] \\$
$\Rightarrow \dfrac{1}{{4\pi {\varepsilon _0}k}}\left[ 0 \right] \\$
$\therefore {V_b} = 0 \\$
Similarly, \[{V_d} = 0 \]
Thus, electrostatic potential is zero at points \[b \] and \[d \] .