Q130 / 180 · NEET 2026 (Cancelled) · Physics
Consider two uncharged capacitors of equal capacitance $200$ pF. One of them is charged by a $100$ V supply and disconnected. Now this capacitor is connected to the uncharged capacitor. The amount of electrostatic energy lost in the process is :
- Correct answerA$0.5 \times 10^{-6}$ J
- B$1.0$ J
- C$1.0 \times 10^{-6}$ J
- D$0.5$ J
Solution
Initial energy of the charged capacitor is given by $U_i = \dfrac{1}{2}CV^2$.
When it is connected to an identical uncharged capacitor, the common potential becomes $V' = \dfrac{CV}{C+C} = \dfrac{V}{2}$.
Final energy of the system is $U_f = \dfrac{1}{2}(2C)V'^2 = \dfrac{1}{2}(2C)\left(\dfrac{V}{2}\right)^2 = \dfrac{1}{4}CV^2$.
Energy lost during the process is $\Delta U = U_i - U_f = \dfrac{1}{2}CV^2 - \dfrac{1}{4}CV^2 = \dfrac{1}{4}CV^2$.
Substituting the given values $C = 200 \times 10^{-12}$ F and $V = 100$ V:
$\Delta U = \dfrac{1}{4} \times 200 \times 10^{-12} \times (100)^2$
$\Delta U = 50 \times 10^{-12} \times 10^4 = 0.5 \times 10^{-6}$ J.
Answer: $0.5 \times 10^{-6}$ J
When it is connected to an identical uncharged capacitor, the common potential becomes $V' = \dfrac{CV}{C+C} = \dfrac{V}{2}$.
Final energy of the system is $U_f = \dfrac{1}{2}(2C)V'^2 = \dfrac{1}{2}(2C)\left(\dfrac{V}{2}\right)^2 = \dfrac{1}{4}CV^2$.
Energy lost during the process is $\Delta U = U_i - U_f = \dfrac{1}{2}CV^2 - \dfrac{1}{4}CV^2 = \dfrac{1}{4}CV^2$.
Substituting the given values $C = 200 \times 10^{-12}$ F and $V = 100$ V:
$\Delta U = \dfrac{1}{4} \times 200 \times 10^{-12} \times (100)^2$
$\Delta U = 50 \times 10^{-12} \times 10^4 = 0.5 \times 10^{-6}$ J.
Answer: $0.5 \times 10^{-6}$ J
Hence, the correct answer is option: 'A'.