In this post, we will solve a numerical problem on the dielectric strength and dielectric constant of capacitor, maximum electric field, maximum potential difference, maximum charge, etc.

## Numerical problem on dielectric strength & dielectric constant – solved

Question: **A parallel-plate capacitor has plates of dimensions 2.0 cm by 3.0 cm separated by a 1.0-mm thickness of paper.****(A) Find its capacitance.****(B) What is the maximum charge that can be placed on the capacitor?**

**Dielectric constant of paper = K = 3.7**

**Dielectric strength of paper = 16 x 10 ^{6} V/m.**

**Solution:**

(A)

K = 3.7

Area =A= 2×3 cm^{2} = 6 x 10^{-4} m^{2}

d=1 mm = 10^{-3} m

ε_{0} = 8.85 x 10 ^{-12} C^{2} / Nm^{2}

Capacitance C = K Aε_{0} / d = 3.7 x 6 x 10^{-4} x 8.85 x 10 ^{-12} /10^{-3} F = 20 x 10^{-12} F = 20 pF

(B) Dielectric strength of paper = 16 x 10^{6} V/m

Hence, maximum electric field = E_{max} = 16 x 10^{6} V/m

Max Potential difference V_{max} = E_{max} . d = 16 x 10^{6} x 10^{-3} V = 16 x 10^{3} V

Hence, here the Maximum Charge = Q_{max} = C V_{max} = 20 x 10^{-12} x 16 x 10^{3} C = 0.32 μC