Question

For inside points $(r \le R)$ $E = 0\, \, \Rightarrow\, \, V =constant = \frac{1}{4 \pi\varepsilon_0} \frac{q}{R}$ For inside points $(r \ge R)$ \hspace18mm E = \frac{1}{4 \pi\varepsilon_0}.\frac{q}{r^2} or $E \propto \frac{1}{r^2}$ and \hspace10mm V = \frac{1}{4 \pi\varepsilon_0}\frac{q}{r} or $V \propto \frac{1}{r}$ On the surface (r = R) \hspace20mm V = \frac{1}{4 \pi\varepsilon_0}\frac{q}{R} \Rightarrow\hspace15mm E = \frac{1}{4 \pi\varepsilon_0}.\frac{q}{R^2}=\frac{\sigma}{\varepsilon_0} where, $\sigma = \frac{q}{4 \pi R^2}=$ surface charge density corresponding to above equations the correct graphs are shown in option (d).

• A. positive
• B. negative
• C. zero
• D. depends on the path connecting the initial and final positions

Answer 1

Answer: (C)

zero

Answer 2

$A \equiv (-a,0,0), B \equiv (0,a,0)$
Point charge is moved from A to B
\hspace25mm V_A=V_B=0
$\therefore$ \hspace15mm W = 0
or the correct option is (c).