Answer to Question #329532 in Electricity and Magnetism for Terdo

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 X 106 V/m(a) What is the potential difference between the plates? (b) What is the area of

each plate? (c) What is the capacitance?

Answer 1

a) The potential difference is:

$$V = Ed$$

where $E= 4.00 \times 10^6 V/m$ and $d=2.50mm = 2.50\times 10^{-3}m$. Thus, obtain:

$$V = 4.00 \times 10^6 V/m\cdot 2.50\times 10^{-3}m = 1.00\times 10^4V$$

c) The capacitance is given as follows:

$$C = \dfrac{q}{V}$$

where $q = 80.0\times 10^{-9}C$. Obtain:

$$C = \dfrac{80.0\times 10^{-9}C}{1.00\times 10^4V}=8.00\times 10^{-13} F$$

b) On the other hand the capacitance of a parallel plate capacitor is:

$$C = \dfrac{\varepsilon_0 A}{d}$$

where $\varepsilon_0\approx 8.85\times 10^{-12} F/m$ is the vacuum permittivity and $A$ is the are of the plate. Expressing $A$ and substituting $C$, obtain:

$$A = \dfrac{dC}{\varepsilon_0} = \dfrac{2.50\times 10^{-3}m\cdot 8.00\times 10^{-13} F}{8.85\times 10^{-12} F/m} \approx 2.26\times 10^{-4}m^2$$