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.
You need to login to perform this action.
You will be redirected in
3 sec