Question

1. Calculate the capacitance of a capacitor whose plates are 25cm x 4.0 cm and are separated by a 2.0mm air gap.
2. What is the charge on each plate if the capacitor is connected to a 12.0 V battery?
3. What is the electric field between the plates?

Answer 1

(1 )

$C=\frac{A\epsilon_0}{d}$

$$C=\frac{10^{-2}\times8.85\times10^{-12}}{2\times10^{-3}}=4.425\times10^{-11}F$$

2

Charge

$$Q=CV=12\times4.425\times10^{-11}=5.31\times10^{-10}C$$

3

$E=\frac{V}{l}$

$E=\frac{12}{0.002}=6000V/m$