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NVAEasyJEE 2025Capacitors & Dielectrics

JEE Physics 2025 Question with Solution

A parallel plate capacitor has charge 5×10−6 C5 \times 10^{-6} \, C. A dielectric slab is inserted between the plates and almost fills the space between the plates. If the induced charge on one face of the slab is 4×10−6 C4 \times 10^{-6} \, C then the dielectric constant of the slab is _____.

Answer

Correct answer:5

Step-by-step solution

Standard Method

Given: Free charge on the capacitor is Q=5×10−6 CQ = 5 \times 10^{-6} \, \text{C} and induced charge on one face of the dielectric slab is Qi=4×10−6 CQ_i = 4 \times 10^{-6} \, \text{C}.

Find: The dielectric constant KK of the slab.

When the dielectric slab is inserted, the induced charge is related to the free charge by

Qi=Q(1−1K)Q_i = Q\left(1 - \frac{1}{K}\right)

Substituting the given values,

4×10−6=5×10−6(1−1K)4 \times 10^{-6} = 5 \times 10^{-6}\left(1 - \frac{1}{K}\right)

Divide both sides by 5×10−65 \times 10^{-6}:

45=1−1K\frac{4}{5} = 1 - \frac{1}{K}

Therefore,

1K=1−45=15\frac{1}{K} = 1 - \frac{4}{5} = \frac{1}{5}

Hence,

K=5K = 5

Therefore, the dielectric constant of the slab is 55.

Using effective charge relation

Given: Q=5×10−6 CQ = 5 \times 10^{-6} \, \text{C} and Qi=4×10−6 CQ_i = 4 \times 10^{-6} \, \text{C}.

Find: The dielectric constant kk.

The effective charge contributing to the electric field inside the capacitor is Q−QiQ - Q_i. For a dielectric medium,

Q=k(Q−Qi)Q = k(Q - Q_i)

So,

k=QQ−Qik = \frac{Q}{Q - Q_i}

Substitute the values:

k=5×10−65×10−6−4×10−6k = \frac{5 \times 10^{-6}}{5 \times 10^{-6} - 4 \times 10^{-6}} k=5×10−61×10−6=5k = \frac{5 \times 10^{-6}}{1 \times 10^{-6}} = 5

Thus, the dielectric constant is 55.

Common mistakes

  • Using the induced charge QiQ_i as the total plate charge is incorrect, because QiQ_i is the polarization charge on the dielectric surface, not the free charge on the capacitor plates. Use Q=5×10−6 CQ = 5 \times 10^{-6} \, \text{C} as the plate charge and relate it properly to QiQ_i.

  • Writing the relation as Qi=QKQ_i = \frac{Q}{K} is wrong, because the induced charge is not equal to the reduced free charge directly. The correct relation is Qi=Q(1−1K)Q_i = Q\left(1 - \frac{1}{K}\right).

  • Subtracting incorrectly in Q−QiQ - Q_i can lead to a wrong value of KK. Here, 5×10−6−4×10−6=1×10−65 \times 10^{-6} - 4 \times 10^{-6} = 1 \times 10^{-6}, not 10−610^{-6} with a changed coefficient or sign. Keep the powers of ten consistent.

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