question_answer

Three identical large metal plates each of area S are at distance d and 2d from each other as shown. Metal plate A is uncharged, while metal plates B and C have charges \[+Q\] and \[-Q\] respectively. Metal plates A and C are connected by a conducting wire through a switch K. How much electrostatic energy is lost when the switch is closed?

A) \[\frac{5{{Q}^{2}}d}{6{{\varepsilon }_{0}}S}\]           

B)        \[\frac{{{Q}^{2}}d}{3{{\varepsilon }_{0}}S}\]

C) \[\frac{{{Q}^{2}}d}{6{{\varepsilon }_{0}}S}\]

D)        \[\frac{2{{Q}^{2}}d}{3{{\varepsilon }_{0}}S}\]

Correct Answer: C

Solution :

[c] Capacitance between A and B is \[C=\frac{{{\varepsilon }_{0}}S}{2d}\] Capacitance between B and C is \[\frac{{{\varepsilon }_{0}}S}{d}=2C\] Initial energy stored \[{{U}_{i}}=\frac{{{Q}^{2}}}{2(2C)}=\frac{{{Q}^{2}}}{4C}\] After the switch is closed, we have two capacitors in parallel, as shown in Figure                         \[{{q}_{1}}+{{q}_{2}}=C\] and       \[\frac{{{q}_{1}}}{C}=\frac{{{q}_{2}}}{2C}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,{{q}_{1}}=\frac{{{q}_{2}}}{2}\] Solving (i) and (ii) we get \[{{q}_{1}}=\frac{Q}{3};\,\,{{q}_{2}}=\frac{2Q}{3}\] Final energy stored in the system \[{{U}_{f}}=\frac{1}{2}\frac{q_{1}^{2}}{c}+\frac{1}{2}\frac{q_{2}^{2}}{2C}=\frac{{{Q}^{2}}}{6C}\] \[\therefore \]  Loss in energy \[\Delta U={{U}_{i}}-{{U}_{f}}\] \[=\frac{{{Q}^{2}}}{4C}-\frac{{{Q}^{2}}}{6C}=\frac{{{Q}^{2}}}{16C}=\frac{{{Q}^{2}}}{12\frac{{{\varepsilon }_{0}}S}{2d}}=\frac{{{Q}^{2}}.d}{6{{\varepsilon }_{0}}.S}\]