Question 151:
easyA parallel plate capacitor has a uniform electric field $E$ in the space between the plates. If the distance between the plates is $d$ and the area of each plate is $A$, the energy stored in the capacitor is: (2021, 2012 Mains, 2011 pre, 2008)
Energy density $u = \frac{1}{2} \varepsilon_{0} E^{2}$. Total energy $U = u \times \text{volume} = \frac{1}{2} \varepsilon_{0} E^{2} A d$.
Question 152:
easyA capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of resulting system: (2017-Delhi)
Initial energy $U_{i} = \frac{Q^{2}}{2C}$. When connected in parallel to an identical capacitor, common potential $V = \frac{Q}{2C}$. Final energy $U_{f} = \frac{1}{2}(2C)V^{2} = \frac{1}{2}(2C)(\frac{Q}{2C})^{2} = \frac{Q^{2}}{4C} = \frac{U_{i}}{2}$. Therefore, the energy decreases by a factor of 2.
Question 153:
easyEnergy per unit volume for a capacitor having area $A$ and separation $d$ kept at potential difference $V$ is given by: (2001)
Energy density $u = \frac{1}{2} \varepsilon_{0} E^{2}$. For a parallel plate capacitor, electric field $E = \frac{V}{d}$. Therefore, $u = \frac{1}{2} \varepsilon_{0} \frac{V^{2}}{d^{2}}$.
Question 154:
easyA capacitor is charged with a battery and energy stored is $U$. After disconnecting battery another capacitor of same capacity is connected in parallel with it. Then energy stored in each capacitor is: (2000)
Initial energy $U = \frac{1}{2}CV^{2}$. Common potential after sharing charge is $V' = \frac{CV}{C+C} = \frac{V}{2}$. Energy in each capacitor is $U' = \frac{1}{2}C(V')^{2} = \frac{1}{2}C(\frac{V}{2})^{2} = \frac{1}{4}(\frac{1}{2}CV^{2}) = \frac{U}{4}$.
Question 155:
easyThe energy stored and potential $V$ is in a capacitor of capacity $C$ given by (1996)
The work done in charging a capacitor is stored as electrostatic potential energy. $U = \int V dq = \int \frac{q}{C} dq = \frac{Q^{2}}{2C}$. Since $Q = CV$, $U = \frac{1}{2}CV^{2} = \frac{C V^{2}}{2}$.