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JEE Main 2025 (7 Apr, Shift 2) · Dielectrics

Dielectrics question from JEE Main 2025 (7 Apr, Shift 2)

Q30JEE Main 2025 · 7 APR · SHIFT 2Easy

Dielectrics

A parallel plate capacitor has charge 5×10−6 C5 \times 10^{-6}\text{ 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}\text{ C} then the dielectric constant of the slab is _______________.

Numerical answer type. Work it out, then open the answer.

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Answer: 5

  1. Induced charge: qb=qf(1−1K)q_b = q_f\left(1 - \frac{1}{K}\right).
  2. 4×10−6=5×10−6(1−1K)⇒1K=0.24 \times 10^{-6} = 5 \times 10^{-6}\left(1 - \frac{1}{K}\right) \Rightarrow \frac{1}{K} = 0.2.
  3. So K=5K = 5.
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