Question

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

• A. $\frac{\sigma \mathrm{RQ}}{\varepsilon_0}\left(\frac{2}{\pi}-\frac{1}{2}\right)$
• B. $\frac{\sigma \mathrm{RQ}}{\varepsilon_0}\left(\frac{1}{2}-\frac{1}{\pi}\right)$
• C. $\frac{\sigma \mathrm{aQ}}{\varepsilon_0}\left(\frac{1}{\pi}-\frac{1}{2}\right)$
• D. $\frac{\sigma \mathrm{aQ}}{\varepsilon_0}\left(\frac{1}{2}-\frac{2}{\pi}\right)$

Answer 1

Answer: (B)

$2 R cos \theta = r$
$d r = - 2 R sin \theta d \theta $
when $r = 0 \, \, \theta = 90^{o} = \frac{\pi }{2}$
$r = 2 R \, \, \theta = 0$

$d V = \frac{K d \theta }{r}$
$d V = \frac{K \left(\right. \sigma r 2 \theta \left.\right) d r}{r}$
$d V = 2 K \sigma \theta d r$
$\displaystyle \int d V = \displaystyle \int 2 K \sigma \theta \left(\right. - 2 R \, sin \theta d \theta \left.\right)$
$V = - 4 K \sigma R \displaystyle \int _{\frac{\pi }{2}}^{0} \theta sin \theta d \theta $
$V=4K\sigma R\displaystyle \int _{0}^{\frac{\pi }{2}} \theta \, sin \theta d \theta $
$V = 4 K \sigma R \left[\right. - \theta cos \theta - \displaystyle \int 1 \left(\right. - cos \theta \left.\right) d \theta \left]\right.$
$V = 4 K \sigma R \left[\right. - \theta cos \theta + sin \theta \left]\right._{0}^{\frac{\pi }{2}}$
$V = 4 K \sigma R = \frac{\sigma }{\pi \epsilon _{0}} R$
For Rim point $V_{R} = V = \frac{\sigma R}{\pi \epsilon _{0}}$
$W = Q \left(\right. V_{f} - V_{i} \left.\right)$
$W = Q \left(\right. V_{C} = V_{R} \left.\right)$
$W=Q(\frac{\sigma R}{2 \varepsilon_0}-\frac{\sigma R}{\pi \varepsilon_0}) $
$ W=\frac{Q \sigma R}{\varepsilon_0}\left(\frac{1}{2}-\frac{1}{\pi}\right)$