Question

Four points $a$, $b$, $c$ and $d$ are set at equal distance from the centre of a dipole as shown in the figure.

The magnitudes of 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

Answer: (B)

Here distance between a and $+ q =$ distance between $C$ and $- q = y _{1}$ (say);
distance between a and $- q =$ distance between $C$ and $+ q = y _{2}$ similarly, $d (+ q )= d (- q )= b (- q )= b (+ q )= r$ (say)
Thus, $V _{ a }=\frac{ kq }{ y _{1}}+\frac{- kq }{ y _{2}}$
$V _{ b }=\frac{ kq }{ r }+\frac{- kq }{ r }=0$
$V _{ c }=\frac{ kq }{ y _{2}}+\frac{- kq }{ y _{1}}$ $V _{ d }=\frac{ kq }{ r }+\frac{- kq }{ r }=0$
Since $y _{2}> y _{1}, V _{ a }$ is positive $V _{ c }$ is negative.
Thus $V _{ a }> V _{ b }= V _{ d }> V _{ c }$