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 $V_a>V_b=V_d>V_c$