Question

A capacitor having a capacitance of 100 µF is charged to a potential difference of 50 V. (a) What is the magnitude of the charge on each plate? (b) The charging battery is disconnected and a dielectric of dielectric constant 2⋅5 is inserted. Calculate the new potential difference between the plates. (c) What charge would have produced this potential difference in absence of the dielectric slab. (d) Find the charge induced at a surface of the dielectric slab.

Answer 1

(a) The magnitude of the charge can be calculated as:
Charge = Capacitance × Potential difference

$\Rightarrow Q=100\times {10}^{-6}\times 50=5\mathrm{mC}$

(b) When a dielectric is introduced, the potential difference decreases.

We know,

$$\begin{gathered} V=\frac{\mathrm{Initial}\mathrm{potential}}{\mathrm{Dielectric}\mathrm{constant}} \\ \Rightarrow V=\frac{50}{2.5}=20\mathrm{V} \end{gathered}$$

(c) Now, the charge on the capacitance can be calculated as:
Charge = Capacitance × Potential difference
$\Rightarrow q_f=20\times 100\times {10}^{-6}=2\mathrm{mC}$

(d) The charge induced on the dielectric can be calculated as:

$q=q_i\left(1-\frac{1}{K}\right)=5\mathrm{mC}\left(1-\frac{1}{2.5}\right)=3\mathrm{mC}$