question_answer

A non-conducting disc of radius R is uniformly charged with surface charge density \[\sigma \]. A disc of radius \[\frac{R}{2}\] is cut from the disc as shown in the figure. The electric potential at centre C of large disc will be

A) \[\frac{\pi \sigma R}{2{{\varepsilon }_{0}}}\]      

B)        \[\frac{\sigma R}{2\pi {{\varepsilon }_{0}}}\]

C) \[\frac{\sigma R(\pi -1)}{2\pi {{\varepsilon }_{0}}}\]        

D)        \[\frac{\sigma R(\pi -1)}{2{{\varepsilon }_{0}}}\]

Correct Answer: B

Solution :

[b] Electric potential at the centre of complete disc, \[{{V}_{1}}=\frac{\sigma R}{2{{\varepsilon }_{0}}}\] Potential at circumference of smaller disc, \[{{V}_{2}}=\frac{\sigma R}{2\pi {{\varepsilon }_{0}}}\] \[\therefore \]  Net potential at centre C is, \[V=({{V}_{1}}-{{V}_{2}})=\frac{\sigma R(\pi -1)}{2\pi {{\varepsilon }_{0}}}\] \[{{V}_{1}}=\int\limits_{r=0}^{R}{\frac{2\pi rdr\times \sigma }{4\pi {{\varepsilon }_{0}}r}}=\frac{\sigma R}{2{{\varepsilon }_{0}}}\] For \[{{V}_{2}}:\] \[r=\frac{2R}{2}\cos \left( \frac{\theta }{2} \right)\] \[dr=-\frac{2R}{2}\sin \left( \frac{\theta }{2} \right)\left( \frac{d\theta }{2} \right)\] \[\int{d{{V}_{2}}}=\int\limits_{\theta =0}^{\pi }{\frac{\sigma \times (r\theta )\,(dr)}{4\pi {{\varepsilon }_{0}}r}}=\frac{\sigma R}{2\pi {{\varepsilon }_{0}}}={{V}_{2}}\]