A parallel plate capacitor having capacitance 12 pF is charged by a battery to a potential difference of 10 V between its plates. The charging battery is now disconnected and a porcelain slab of dielectric constant 6.5 is slipped between the plates. The work done by the capacitor on the slab is :

Capacitor PYQ — JEE Main 2019 · 10 Jan · Shift 2 · Q56 with Solution

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Correct answer: A

Given data
- Initial capacitance: $C_0 = 12\,\text{pF} = 12 \times 10^{-12}\,\text{F}$
- Initial potential difference: $V_0 = 10\,\text{V}$
- Dielectric constant of porcelain: $K = 6.5$
- Battery is disconnected, so charge remains constant.
Initial charge on the capacitor

Since the battery is disconnected after charging, $Q = C_0 V_0 = (12 \times 10^{-12})(10) = 120 \times 10^{-12}\,\text{C}$
Initial energy stored

$U_i = \frac{1}{2} C_0 V_0^2$ $U_i = \frac{1}{2}(12 \times 10^{-12})(10)^2$ $U_i = \frac{1}{2}(12 \times 10^{-12})(100)$ $U_i = 600 \times 10^{-12}\,\text{J} = 600\,\text{pJ}$
New capacitance after inserting dielectric

When the dielectric fully fills the space between the plates, $C' = K C_0 = 6.5 \times 12\,\text{pF} = 78\,\text{pF}$
Final energy stored

Since charge remains constant, use $U_f = \frac{Q^2}{2C'}$

But $Q^2/(2C_0) = U_i$ , so $U_f = \frac{U_i}{K} = \frac{600}{6.5}\,\text{pJ}$ $U_f \approx 92.31\,\text{pJ}$
Work done by the capacitor on the slab

The electrostatic energy decreases, and that decrease appears as work done on the slab: $W = U_i - U_f$ $W = 600 - 92.31 = 507.69\,\text{pJ}$

$W \approx 508\,\text{pJ}$
Option check
- A: $508\,\text{pJ}$ ✅
- B: $692\,\text{pJ}$ ❌
- C: $560\,\text{pJ}$ ❌
- D: $600\,\text{pJ}$ ❌

Therefore, the correct answer is A.