JEE Main 2025 Apr 7 Shift 2, Physics Q21: Capacitance For Geometrical Figures With Dielectric

JEE Main2025Apr 7, Shift 2Physics

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 _______.

Your answer

Solution

When a dielectric of dielectric constant $K$ is inserted almost completely between the plates of a charged capacitor with free charge $Q$ , the bound (induced) charge on each face of the slab is $Q_{in d} = Q (1 - \frac{1}{K})$ .

Substituting the given values: $4 \times 1 0^{- 6} = 5 \times 1 0^{- 6} (1 - \frac{1}{K}) \Rightarrow 1 - \frac{1}{K} = \frac{4}{5} \Rightarrow \frac{1}{K} = \frac{1}{5}$ .

$K = 5$ .

Concept behind this question

Capacitance For Geometrical Figures With DielectricNotes, formulas and examples →

Practice more Physics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →

JEE Main 2025 Apr 7 Shift 2, Physics Q21: Capacitance For Geometrical Figures With Dielectric

Concept behind this question

More previous year questions on Capacitance For Geometrical Figures With Dielectric

Practice more Physics