​Two parallel plates (each of area 200 $$\text{cm}^2$$​) of a capacitor are given equal and opposite charges of magnitude 1.78 μC. The space between the plates is filled with a dielectric. If the electric field in the dielectric is 1.5 × $$10^6$$​ V/m, what will be the magnitude of charge induced on each dielectric plate?

$$\left(\varepsilon_0=8.85 \times 10^{-12} \mathrm{~F} / \mathrm{m}\right)$$​​

Correct option is A

​$$\begin{aligned}&\text{Given:} \\&\bullet \ \text{Plate area } A = 200 \, \text{cm}^2 = 200 \times 10^{-4} \, \text{m}^2 \\&\bullet \ \text{Electric field in dielectric } E = 1.5 \times 10^6 \, \text{V/m} \\&\bullet \ \text{Free charge } Q = 1.78 \times 10^{-6} \, \text{C} \\&\bullet \ \text{Vacuum permittivity } \varepsilon_0 = 8.85 \times 10^{-12} \, \text{F/m} \\[10pt]&\text{Induced charge on dielectric is given by:} \\&Q' = \left(1 - \frac{1}{k} \right) Q \\[6pt]&\text{where } k \text{ is the dielectric constant.} \\[6pt]&\text{Also, dielectric constant:} \\&k = \frac{Q}{\varepsilon_0 E A}\end{aligned}$$

$$\begin{aligned}&1. \ \text{dielectric constant:} \\&k = \frac{1.78 \times 10^{-6}}{8.85 \times 10^{-12} \times 1.5 \times 10^6 \times 200 \times 10^{-4}} = 6.70 \\[10pt]&2. \ \text{induced charge:} \\&Q' = \left(1 - \frac{1}{6.70} \right) \times 1.78 = 1.51 \, \mu C\end{aligned}$$​​​