question_answer

Two plates (area = S) charged to +\[{{q}_{1,}}\] and +\[{{q}_{2,}}\]\[({{q}_{2}}<{{q}_{2}})\]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 :

\[V=\frac{({{q}_{1}}-{{q}_{2}})d}{2{{\varepsilon }_{0}}S}\] \[=\frac{({{q}_{1}}-{{q}_{2}})}{2{{\varepsilon }_{0}}\frac{S}{d}}\]                 \[=\frac{{{q}_{1}}-{{q}_{2}}}{2C}\]