Question

A parallel plate capacitor has a capacitance of 120 pF (without dielectric) and plate area of 100 cm$^2$. The space between the plates is now filled with Mica ($k = 5.0$). The potential difference between the plates is 50 V. What is modulus of charge induced on either surface of mica (in nC)?

Answer 1

24 nC

Answer 2

The capacitance of the parallel plate capacitor without dielectric is given as $C_0 = 120 \text{ pF}$.

The dielectric constant of Mica is $K = 5.0$.

The potential difference between the plates is $V = 50 \text{ V}$.

When the space between the plates is filled with a dielectric of constant $K$, the capacitance becomes $C = K C_0$.

$C = 5.0 \times 120 \text{ pF} = 600 \text{ pF}$.

The free charge on the plates of the capacitor when the potential difference $V$ is applied is given by $Q_{free} = C V$.

$Q_{free} = 600 \text{ pF} \times 50 \text{ V}$

$Q_{free} = (600 \times 10^{-12} \text{ F}) \times 50 \text{ V}$

$Q_{free} = 30000 \times 10^{-12} \text{ C}$

$Q_{free} = 3 \times 10^{-8} \text{ C}$

$Q_{free} = 30 \times 10^{-9} \text{ C} = 30 \text{ nC}$.

The modulus of the charge induced on either surface of the dielectric is given by the formula:

$|Q_{induced}| = Q_{free} \left(1 - \frac{1}{K}\right)$

Substituting the values of $Q_{free}$ and $K$:

$|Q_{induced}| = 30 \text{ nC} \left(1 - \frac{1}{5.0}\right)$

$|Q_{induced}| = 30 \text{ nC} \left(1 - 0.2\right)$

$|Q_{induced}| = 30 \text{ nC} \times 0.8$

$|Q_{induced}| = 24 \text{ nC}$.

The modulus of the charge induced on either surface of mica is 24 nC.