PROPERTIES OF A PARALLEL-PLATE CAPACITOR
The plates of a parallel-plate capacitor in vacuum are 5.00 mm apart and 2.00 m^2 in area. A 10.0-kV potential difference is applied across the capacitor. Compute (a) the capacitance; (b) the charge on each plate; and (c) the magnitude of the electric field between the plates.
IDENTIFY and SET UP:
We are given the plate area A, the plate spacing d, and the potential difference V_{ab} = 1.00 × 10^4 V for this parallel-plate capacitor. Our target variables are the capacitance C, the charge Q on each plate, and the electric-field magnitude E.
We use Eq. (24.2) to calculate C and then use Eq. (24.1) and V_{ab} to find Q. We use E=Q / \epsilon_{0} A to find E.
C=\frac{Q}{V_{a b}} (24.1)
C=\frac{Q}{V_{a b}}=\epsilon_{0} \frac{A}{d} (24.2)
EXECUTE:
(a) From Eq. (24.2),
\begin{aligned}C &=\epsilon_{0} \frac{A}{d}=\left(8.85 \times 10^{-12} \mathrm{~F} /\mathrm{m}\right) \frac{\left(2.00 \mathrm{~m}^{2}\right)}{5.00 \times 10^{-3}\mathrm{~m}}\\&=3.54 \times 10^{-9} \mathrm{~F}=0.00354 \mu \mathrm{F}\end{aligned}(b) The charge on the capacitor is
\begin{aligned}Q=C V_{a b} &=\left(3.54 \times 10^{-9} \mathrm{C} / \mathrm{V}\right)\left(1.00 \times 10^{4} \mathrm{~V}\right) \\&=3.54 \times 10^{-5} \mathrm{C}=35.4 \mu\mathrm{C}\end{aligned}The plate at higher potential has charge +35.4 μC, and the other plate has charge -35.4 μC.
(c) The electric-field magnitude is
\begin{aligned}E &=\frac{\sigma}{\epsilon_{0}}=\frac{Q}{\epsilon_{0} A}=\frac{3.54 \times10^{-5} \mathrm{C}}{\left(8.85 \times 10^{-12} \mathrm{C}^{2} / \mathrm{N} \cdot\mathrm{m}^{2}\right)\left(2.00 \mathrm{~m}^{2}\right)} \\&=2.00 \times 10^{6}\mathrm{~N} / \mathrm{C}\end{aligned}
EVALUATE: We can also find E by recalling that the electric field is equal in magnitude to the potential gradient [Eq. (23.22), \vec{E}=-\vec{\nabla} V]. The field between the plates is uniform, so
E=\frac{V_{a b}}{d}=\frac{1.00 \times 10^{4} \mathrm{~V}}{5.00 \times 10^{-3} \mathrm{~m}}=2.00 \times 10^{6} \mathrm{~V} / \mathrm{m}(Remember that 1 N/C = 1 V/m.)