Question

The plates of a parallel-plate capacitor are 2.50 mm apart, and each carries a charge of magnitude 80.0 nC. The plates are in vacuum. The electric field between the plates has a magnitude of 4.00×106V/m. What is (a) the potential difference between the plates; (b) the area of each plate; (c) the capacitance?

Answer 1

Answer:

10000 V

0.00225988700565 m²

$8\times 10^{-12}\ F$

Explanation:

E = Electric field = $4\times 10^6\ V/m$

d = Gap = 2.5 mm

Q = Charge = 80 nC

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

Potential difference is given by

$V=Ed\\\Rightarrow V=4\times 10^6\times 2.5\times 10^{-3}\\\Rightarrow V=10000\ V$

The potential difference between the plates is 10000 V

Area is given by

$A=\dfrac{Q}{\epsilon_0E}\\\Rightarrow A=\dfrac{80\times 10^{-9}}{8.85\times 10^{-12}\times 4\times 10^6}\\\Rightarrow A=0.00225988700565\ m^2$

The area of the plate is 0.00225988700565 m²

Capacitance is given by

$C=\dfrac{\epsilon_0A}{d}\\\Rightarrow C=\dfrac{8.85\times 10^{-12}\times 0.00225988700565}{2.5\times 10^{-3}}\\\Rightarrow C=8\times 10^{-12}\ F$

The capacitance is $8\times 10^{-12}\ F$