Question

A 3.0 cm × 3.0 cm parallel-plate capacitor has a 3.0 mm spacing. The electric field strength inside the capacitor is 1.2×105 V/m . What is the potential difference across the capacitor? How much charge is on each plate?

Answer 1

To solve this problem it is necessary to apply the concepts related to the voltage depending on the electric field and the distance, as well as the load depending on the capacitance and the voltage. For the first part we will use the first mentioned relationship, for the second part, we will not only define the load as the capacitance by the voltage but also place it in terms of the Area, the permittivity in free space, the voltage and the distance.

PART A ) Voltage in function of electric field and distance can be defined as,

$V = Ed$

Our values are,

$E = 1.2*10^5 V/m$

$d = 3.0mm = 3*10^{-3}$

Replacing,

$V = (1.2*10^5)(3*10^{-3})$

$V = 360v$

Therefore the potential difference across the capacitor is 360V

PART B) The charge can be defined as,

$Q = CV = \frac{\epsilon AV}{d}$

Here,

$\epsilon = 8.85*10^{-12} F/m$, Permittivity of free space

$A = s^2$, area of each capacitor plate

s = Length of capacitor plate

Replacing,

$Q = \frac{\epsilon AV}{d}$

$Q = \frac{(8.85*10^{-12})(0.03)^2(240)}{2.0*10^{-8}m}$

$Q = 9.558*10^{-10}C$

Therefore the charge on each plate is $9.558*10^{-10}C$