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 \[\Delta U={{U}_{i}}-{{U}_{f}}\] \[=\frac{{{Q}^{2}}}{4C}-\frac{{{Q}^{2}}}{6C}=\frac{{{Q}^{2}}}{16C}=\frac{{{Q}^{2}}}{12\frac{{{\varepsilon }_{0}}S}{2d}}=\frac{{{Q}^{2}}.d}{6{{\varepsilon }_{0}}.S}\]You need to login to perform this action.
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