Question

Work done by an external agent to move slowly a charge $Q$ from rim of a uniformly charged horizontal disc of radius $R$ and charge per unit area $\sigma$ , to center of this disc is -

• A. $\frac{\sigma RQ}{{\epsilon }_0}\left(\frac{2}{\pi }-\frac{1}{2}\right)$
• B. $\frac{\sigma RQ}{{\epsilon }_0}\left(\frac{1}{2}-\frac{1}{\pi }\right)$
• C. $\frac{\sigma RQ}{{\epsilon }_0}\left(\frac{1}{\pi }-\frac{1}{2}\right)$
• D. $\frac{\sigma RQ}{{\epsilon }_0}\left(\frac{1}{2}-\frac{2}{\pi }\right)$

Answer 1

Answer: (B)

$2Rcos\theta =r$
$dr=-2Rsin\theta d\theta$
when $r=0;\theta =90^o=\frac{\pi }{2}$
$r=2R;\theta =0$

$dV=\frac{KdQ}{r}$
$dV=\frac{K(\sigma r2\theta )dr}{r}$
$dV=2K\sigma \theta dr$
$\int dV=\int 2K\sigma \theta (-2Rsin\theta d\theta )$
$V=-4K\sigma R{\int }_{\frac{\pi }{2}}^0\theta sin\theta d\theta$
$V=4K\sigma R{\int }_0^{\frac{\pi }{2}}\theta sin\theta d\theta$
$V=4K\sigma R[-\theta cos\theta -\int 1(-cos\theta )d\theta ]$
$V=4K\sigma R[-\theta cos\theta +sin\theta {]}_0^{\frac{\pi }{2}}$
$V=4K\sigma R=\frac{\sigma }{\pi {ϵ}_0}R$
For Rim point $V_R=V=\frac{\sigma R}{\pi {ϵ}_0}$
$W=Q(V_f-V_i)$
$W=Q(V_C-V_R)$
$W=Q\left(\frac{\sigma R}{2{\epsilon }_0}-\frac{\sigma R}{\pi {\epsilon }_0}\right)$
$W=\frac{Q\sigma R}{{\epsilon }_0}\left(\frac{1}{2}-\frac{1}{\pi }\right)$