Question

Q. 23 a parallel plate capacitor of capacitance 90 pf is connected to a battery of emf 20v. If a dielectric material of dielectric constant is inserted between the plates, the magnitude of the induced charge will be:

Answer 1

you forgot to write value of dielectric constant.
here, I observed , dielectric constant , k = 5/3

first of all, find charge on parallel plate capacitor.
use formula, Q = CV
where C is capacity of capacitance, V is potential difference .

here, C = 90pf and V = 20V
now, Q = 90 × 20 = 1800pC
Q = 1.8 nC

$Q_{\textbf{induced}}=Q_0\left[1-\frac{1}{K}\right]$

or, $Q_{\textbf{induced}}=1.8\left[1-\frac{3}{5}\right]$

or, $Q_{\textbf{induced}}=1.8\left[\frac{2}{5}\right]$

hence, $Q_{\textbf{induced}}=0.72nC$