Question

Derive the expression for the capacitance of a parallel plate capacitor (with and without dielectric slab).

Hint:

Remember: - Capacitance increases with area and dielectric constant. - Capacitance decreases with plate separation. - Dielectric reduces electric field and increases charge storing ability.

Answer 1

Capacitance of a Parallel Plate Capacitor (Without Dielectric)
A parallel plate capacitor consists of two large conducting plates placed parallel to each other and separated by a small distance \(d\). The plates carry equal and opposite charges \(+Q\) and \(-Q\). The region between the plates is usually filled with air or vacuum.

Let the area of each plate be \(A\) and the distance between them be \(d\). The surface charge density on the plates is
\[ \sigma = \frac{Q}{A} \] The electric field between the plates is given by
\[ E = \frac{\sigma}{\varepsilon_0} \] Substituting the value of \( \sigma \):
\[ E = \frac{Q}{A\varepsilon_0} \] The potential difference between the plates is
\[ V = Ed \] Substituting the value of \(E\):
\[ V = \frac{Q}{A\varepsilon_0} \times d \] The capacitance of a capacitor is defined as
\[ C = \frac{Q}{V} \] Substituting the value of \(V\):
\[ C = \frac{Q}{\frac{Qd}{A\varepsilon_0}} \] \[ C = \frac{\varepsilon_0 A}{d} \] Thus, the capacitance of a parallel plate capacitor without dielectric is
\[ C = \frac{\varepsilon_0 A}{d} \] where \( \varepsilon_0 \) is the permittivity of free space.

Capacitance of a Parallel Plate Capacitor (With Dielectric Slab)
When a dielectric material is inserted between the plates, the electric field inside the capacitor decreases due to polarization of the dielectric. Let the dielectric constant of the material be \(K\). The permittivity of the dielectric becomes
\[ \varepsilon = K\varepsilon_0 \] The capacitance of the capacitor with dielectric becomes
\[ C = \frac{\varepsilon A}{d} \] Substituting \( \varepsilon = K\varepsilon_0 \):
\[ C = \frac{K\varepsilon_0 A}{d} \] Thus, the presence of a dielectric increases the capacitance by a factor of \(K\). The dielectric reduces the electric field and allows the capacitor to store more charge for the same potential difference.

Conclusion
For a parallel plate capacitor without dielectric,
\[ C = \frac{\varepsilon_0 A}{d} \] and when a dielectric slab of dielectric constant \(K\) is inserted between the plates, the capacitance becomes
\[ C = \frac{K\varepsilon_0 A}{d} \] Hence, the capacitance increases when a dielectric medium is placed between the plates.