Question

Two plates of area 7.00 × 10-3 m2 are separated by a distance of 2.60 × 10-4 m. If a charge of 5.40 × 10-8 C is moved from one plate to the other, calculate the potential difference (voltage) between the two plates. Assume that the separation distance is small in comparison to the diameter of the plates.

Answer 1

Answer:

226.53 Volt

Explanation:

A = Area of plates = 7.00×10⁻³ m²

ε₀ = Permittivity of space = 8.854×10⁻¹² F/m

d = Distance between two plates = 2.60×10⁻⁴ m

Q = Charge = 5.40×10⁻⁸ C

Capacitance

$C=\frac{\epsilon_0A}{d}\\\Rightarrow C=\frac{8.854\times 10^{-12}\times 7\times 10^{-3}}{2.6\times 10^{-4}}\\\Rightarrow C=23.83\times 10^{-11}$

Potential difference between plates

$V=\frac{Q}{C}\\\Rightarrow V=\frac{5.4\times 10^{-8}}{23.83\times 10^{-11}}\\\Rightarrow V=226.53\ Volt$

∴ The potential difference (voltage) between the two plates is 226.53 Volt