NEET 2022 Physics Energy Stored in Capacitors MCQ Question

Chapter: Electrostatic Potential and Capacitance

Class: 12 Physics Topic: Energy Stored in Capacitors Difficulty: 🟡 Medium

✅ Ans: A — $2.25\times10^{-6}\,J$

Given:

$$C=900\,\text{pF}=900\times10^{-12}\,F$$ $$V=100\,V$$

Initial energy:

$$U_i=\frac12CV^2$$ $$U_i=\frac12(900\times10^{-12})(100)^2$$ $$U_i=4.5\times10^{-6}\,J$$

After connecting an identical uncharged capacitor in parallel:

$$C_{\text{eq}}=2C$$

Charge is conserved, so final voltage becomes:

$$V_f=\frac{100}{2}=50\,V$$

Final energy:

$$U_f=\frac12(2C)(50)^2$$ $$\boxed{U_f=2.25\times10^{-6}\,J}$$

❌ Why other options are wrong?

• A: $2.25\times10^{-6}\,J$ ✅ Correct.
• B: $1.5\times10^{-6}\,J$ ❌ Incorrect.
• C: $4.5\times10^{-6}\,J$ ❌ This is the initial energy.
• D: $3.25\times10^{-6}\,J$ ❌ Incorrect.

📌 NCERT Concept

For identical capacitors, one initially charged and one uncharged:

$$\boxed{V_f=\frac{V}{2}}$$

and

$$\boxed{U_f=\frac{U_i}{2}}$$

🧠 NEET Trick

Identical capacitor + uncharged capacitor → final voltage = half → final energy = half.

$$\boxed{\text{Answer: A}}$$