Question

Obtain the expression for the capacitance of a parallel plate capacitor with a dielectric medium between its plates.

Hint:

When inserting a dielectric material between the plates of a capacitor, the dielectric constant (\(K\)) of the material not only increases the capacitance but also affects the capacitor's ability to store electrical energy without breaking down. This is because the dielectric material reduces the electric field within the capacitor, allowing it to store more charge at the same voltage, or maintain the same charge at a lower voltage.

Answer 1

The capacitance \( C \) of a parallel plate capacitor without any dielectric between its plates is given by: \[ C_0 = \frac{\epsilon_0 A}{d} \] where \( \epsilon_0 \) is the permittivity of free space, \( A \) is the area of one plate, and \( d \) is the separation between the plates. When a dielectric medium with dielectric constant \( K \) is introduced between the plates, the capacitance increases by a factor of \( K \), because the dielectric reduces the effective electric field within the capacitor while maintaining the same charge. Thus, the new capacitance \( C \) is given by: \[ C = KC_0 = K \frac{\epsilon_0 A}{d} \] This expression shows that the capacitance of a parallel plate capacitor is directly proportional to the dielectric constant of the medium between the plates, the permittivity of free space, and the area of the plates, and inversely proportional to the distance between the plates.