Question

A parallel-plate air capacitor is made from two plates 0.210 m square, spaced 0.815 cm apart. it is connected to a 120 v battery. if the plates are pulled apart to a separation of 1.63 cm , suppose the battery remains connected while the plates are pulled apart. part a what is the capacitance?

Answer 1

Answer:

at the beginning: $2.3\cdot 10^{-10} F$

when the plates are pulled apart: $1.1\cdot 10^{-10} F$

Explanation:

The capacitance of a parallel-plate capacitor is given by

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

where

k is the relative permittivity of the medium (for air, k=1, so we can omit it)

$\epsilon_0 = 8.85\cdot 10^{-12} F/m$ is the permittivity of free space

A is the area of the plates of the capacitor

d is the separation between the plates

In this problem, we have:

$A=0.210 m^2$ is the area of the plates

$d=0.815 cm=8.15\cdot 10^{-3} m$ is the separation between the plates at the beginning

Substituting into the formula, we find

$C=(1)(8.85\cdot 10^{-12}F/m)\frac{0.210 m^2}{8.15\cdot 10^{-3} m}=2.3\cdot 10^{-10} F$

Later, the plates are pulled apart to $d=1.63 cm=0.0163 m$, so the capacitance becomes

$C=(1)(8.85\cdot 10^{-12}F/m)\frac{0.210 m^2}{0.0163 m}=1.1\cdot 10^{-10} F$