Question

A parallel-plate capacitor has plates 0.15 mm apart and dielectric with relative permittivity of 3. Find the electric field intensity and the voltage between plates if the surface charge is 5 × 10⁻⁴ μC/cm²

Answer 1

The electric intensity between the plates is

E = D/ε₀ε_r volt/metre;

Now, σ = 5 × 10⁻⁴ μC/cm² = 5 × 10⁻⁶ C/m² 

Since, charge density equals flux density

∴ E = D/ε₀ε_r = (5 x 10^-6)/(8.854 x 10^-12 x 3) = 188, 000V/m = 188KV/m

Now potential difference V = E × dx = 188,000 × (0.15 × 10⁻³ ) = 2.82 V