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 magnitude of electrostatic potential $V_a,V_b$, $V_c$ and $V_d$ would satisfy the following reaction

• 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

The correct option is C $V_a=V_c>V_b=V_d$

$V_a$ due to $q^{+}=\frac{1}{4\pi {\epsilon }_0}\frac{q}{(x-l)}$

$V_a$ due to $q^{-}=\frac{-1}{4\pi {\epsilon }_0}\frac{q}{(x+l)}$

$\therefore v_{atotal}=\frac{q}{4\pi {\epsilon }_0}[\frac{1}{(x-l)}-\frac{1}{(x+l)}]$

$V_a=\frac{q}{4\pi {\epsilon }_0}\frac{2l}{(x^2-l^2)}\Rightarrow V_a=\frac{2lq}{4\pi {\epsilon }_0{x}^2}(ifl<<x)=\frac{p}{4\pi {\in }_0{x}^2}$

$V_c=\frac{q}{4\pi {\epsilon }_0}\{-\frac{1}{(x-l)}+\frac{1}{(x+l)}\}=\frac{-2lq}{4\pi {\epsilon }_0\left(x^2\right)}$

$V_d=\frac{+q}{4\pi {\epsilon }_0}\{\frac{1}{r}-\frac{1}{r}\}$ due to $+q$ and $-q=0$

$V_b=0$

$\therefore V_a=V_c>V_b=V_d$