JEE Main · 2025 · 07 Apr 2025 · Shift 2 · Capacitance

A parallel plate capacitor has charge

5 \times 1 0^{- 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 \times 1 0^{- 6}

C then the dielectric constant of the slab is ______.

Answer: 5

Solution

The charge induced on a face of the dielectric relates to the free charge by

Q_{ind} = Q (1 - \frac{1}{K})

. Substituting

Q = 5 \times 1 0^{- 6} C

and

Q_{ind} = 4 \times 1 0^{- 6} C

gives

\frac{4}{5} = 1 - \frac{1}{K}

, so

\frac{1}{K} = \frac{1}{5}

. The dielectric constant is

5

JEE Main · 2025 · 07 Apr 2025 · Shift 2 · Integer answer

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