question_answer

A parallel plate capacitor is charged to a potential difference V by a DC source, The capacitor is then disconnected from the source. If the distance between the plates is tripled, state with reason, how the following will change?

(i) Charge on capacitor

(ii) Electric potential and field between the plates.

(iii) Energy stored in the capacitor

Answer:

After disconnection from battery and tripling the separation between two plates (i) Charge on capacitor remains same. i.e.        CV = C?V? \[\Rightarrow CV=\left( \frac{C}{3} \right)V'\]                    \[\Rightarrow V'=3\,V\] (ii) Electric field between the plates             \[E'=\frac{V'}{d'}=\frac{3V}{3d}\]             \[E'=\frac{V}{d}=E\] \[\Rightarrow \] Electric field between the two plates remains same. (iii) Capacitance reduces to one third of original value as             \[C\propto \frac{1}{d}\]  \[\Rightarrow C'=\frac{C}{3}\]     Energy stored in the capacitor before disconnection from battery Now, energy stored in the capacitor after disconnection from battery \[{{U}_{2}}=\frac{{{q}^{2}}}{2(C')}=\frac{{{q}^{2}}}{2\times \left( \frac{C}{3} \right)}=\frac{3{{q}^{2}}}{2C}\] \[\Rightarrow \]   \[{{U}_{2}}=\frac{3{{q}^{2}}}{2C}\] \[\Rightarrow \]   \[{{U}_{2}}=3{{U}_{1}}\] Energy stored in capacitor gets .tripled to its initial value.