question_answer

Two plates (area = S) charged to \[+{{q}_{1}}\] and \[+{{q}_{2}}({{q}_{2}}<{{q}_{1}})\] brought closer to form a capacitor of capacitance C. The potential difference across the plates is:

A)  \[\frac{{{q}_{1}}-{{q}_{2}}}{2C}\]

B)  \[\frac{{{q}_{1}}-{{q}_{2}}}{C}\]

C)  \[\frac{{{q}_{1}}-{{q}_{2}}}{4C}\]

D)  \[\frac{2({{q}_{1}}-{{q}_{2}})}{C}\]

Correct Answer: A

Solution :

 Capacitance of the parallel plate capacitor is \[C=\frac{{{\varepsilon }_{0}}\,A}{d}\] where A is area of plates and d is distance between them. Potential difference across the plates \[V=Ed\] Here, \[E=\frac{({{q}_{1}}-{{q}_{2}})}{2\,{{\varepsilon }_{0}}\,A}\] \[\therefore \] \[V=\frac{({{q}_{1}}-{{q}_{2}})\,d}{2\,{{\varepsilon }_{0}}A}=\frac{{{q}_{1}}-{{q}_{2}}}{2C}\].