Answer to Question #159614 in Electric Circuits for Hillary

Question

An air filled capacitor consists of two parallel plates, each with an area of 7.60 cm^2, separated by a distance of 1.80mm. A 20.0V potential difference is applied to these plates. Calculate
(i) the electric field between the plates
(ii) the capacitance and the charge on each plate.

Answer 1

(i) The electric field between the plates can be found as follows:

$$E=\dfrac{V}{d}=\dfrac{20\ V}{1.8\cdot10^{-3}\ m}=11.1\cdot10^3\ \dfrac{V}{m}.$$

(ii) The capacitance of the capacitor can be found as follows:

$$C=\dfrac{\epsilon_0A}{d},$$ $$C=\dfrac{8.85\cdot10^{-12}\ \dfrac{F}{m}\cdot7.6\cdot10^{-4}\ m^2}{1.8\cdot10^{-3}\ m}=3.74\cdot10^{-12}\ F=3.74\ pF.$$

The charge on each plate can be found as follows:

$$Q=C\Delta V,$$ $$Q=3.74\cdot10^{-12}\ F\cdot20\ V=74.8\cdot10^{-12}\ C=74.8\ pC.$$