A parallel plate capacitor of capacitance 90 pF is connected to a battery of EMF 20 V. If a dielectric material of dielectric constant $$K = \frac{5}{3}$$ is inserted between the plates, the magnitude of the induced charge will be:

We have a parallel-plate capacitor whose initial capacitance is

$$C_0 = 90 \,\text{pF} = 90 \times 10^{-12}\,\text{F}.$$

The plates are connected to a battery that keeps a constant potential difference

$$V = 20\,\text{V}.$$

The free charge already present on the plates before any dielectric is introduced is obtained from the basic definition of capacitance

$$C = \dfrac{Q}{V}\; \Longrightarrow\; Q_0 = C_0 V.$$

Substituting the given numbers,

$$Q_0 = (90 \times 10^{-12}\,\text{F})(20\,\text{V}) = 1800 \times 10^{-12}\,\text{C} = 1.8 \times 10^{-9}\,\text{C} = 1.8\,\text{nC}.$$

Now a slab of dielectric having dielectric constant $$K=\dfrac{5}{3}$$ completely fills the space between the plates. As long as the battery remains connected, the voltage stays fixed at $$20\ \text{V}$$, but the capacitance becomes

$$C = K C_0 = \dfrac{5}{3}\,(90 \times 10^{-12}\,\text{F}) = 150 \times 10^{-12}\,\text{F} = 150\,\text{pF}.$$

With this new capacitance, the free charge residing on the metal plates is

$$Q_{\text{free}} = C V = (150 \times 10^{-12}\,\text{F})(20\,\text{V}) = 3000 \times 10^{-12}\,\text{C} = 3.0 \times 10^{-9}\,\text{C} = 3.0\,\text{nC}.$$

The appearance of the dielectric polarises it, creating bound (induced) charges on its two faces. In a uniformly filled capacitor the magnitude of the induced charge is related to the free charge by the well-known relation

$$Q_{\text{ind}} = Q_{\text{free}}\!\left(1 - \dfrac{1}{K}\right).$$

This comes from the vector equation $$\mathbf D = \varepsilon_0\mathbf E + \mathbf P$$ and the fact that the normal component of $$\mathbf D$$ ends on the free charge while the polarisation $$\mathbf P$$ gives the bound charge.

Substituting $$K = \dfrac{5}{3}$$ and $$Q_{\text{free}} = 3.0\,\text{nC}$$, we get

$$Q_{\text{ind}} = 3.0\,\text{nC}\!\left(1 - \dfrac{1}{\,5/3\,}\right) = 3.0\,\text{nC}\!\left(1 - \dfrac{3}{5}\right) = 3.0\,\text{nC}\!\left(\dfrac{2}{5}\right) = 1.2\,\text{nC}.$$

Thus the magnitude of the induced (bound) charge on either face of the dielectric sheet is $$1.2\,\text{nC}$$.

Hence, the correct answer is Option B.