Question

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

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 $V_a>V_b=V_d>V_c$