Question

Derive an expression for the capacitance of a parallel plate capacitor with a dielectric slab between the plates.

Hint:

The insertion of a dielectric material between the plates of a capacitor increases its capacitance by a factor equal to the dielectric constant \( \kappa \).

Answer 1

Step 1: Capacitance of a parallel plate capacitor without dielectric. 
The capacitance of a parallel plate capacitor without a dielectric is given by the formula: \[ C_0 = \frac{\epsilon_0 A}{d} \] where: 
- \( C_0 \) is the capacitance without the dielectric, 
- \( \epsilon_0 \) is the permittivity of free space, 
- \( A \) is the area of the plates, 
- \( d \) is the distance between the plates.
Step 2: Introducing the dielectric. 
When a dielectric material with dielectric constant \( \kappa \) is inserted between the plates, the capacitance increases. The new capacitance \( C \) is given by: \[ C = \kappa C_0 = \kappa \frac{\epsilon_0 A}{d} \] where \( \kappa \) is the dielectric constant of the material, which is greater than 1. 
Step 3: Electric field inside the dielectric. 
When a dielectric is present, the electric field inside the capacitor is reduced by a factor of \( \kappa \). The dielectric reduces the effective electric field between the plates, which leads to an increase in capacitance. This is because the dielectric material polarizes under the influence of the electric field, thereby reducing the effective field inside the capacitor. 
Step 4: Conclusion. 
Thus, the capacitance of a parallel plate capacitor with a dielectric slab between the plates is: \[ C = \kappa \frac{\epsilon_0 A}{d} \]