Answer
(a) $2.64\times 10^{-5}J$
(b) $5.29\times 10^{-5}J$
(c) $2.65\times 10^{-5}J$
Work Step by Step
(a) We can find the required stored energy in the capacitor as follows:
$U=\frac{1}{2}CV^2$
$U=\frac{1}{2}(\frac{K\epsilon_{\circ}A}{d})V^2$
We plug in the known values to obtain:
$U=\frac{1}{2}(\frac{(1.00059)(8.85\times 10^{-12}C^2/N.m^2)(0.0405m^2)}{2.25\times 10^{-3}m})(575V)^2$
$U=2.64\times 10^{-5}J$
(b) We can find the required stored energy in the capacitor as follows:
$U^{\prime}=\frac{1}{2}CV^{\prime 2}$
$U^{\prime}=\frac{1}{2}(\frac{K\epsilon_{\circ}A}{d^{\prime}})V^{\prime 2}$
We know that doubling the separation results in doubled voltage:
$V^{\prime}=2(575V)=1150V$
We plug in the known values to obtain:
$U^{\prime}=\frac{1}{2}(\frac{(1.00059)(8.85\times 10^{-12}C^2/N.m^2)(0.0405m^2)}{4.5\times 10^{-3}m})(1150V)^2$
$U^{\prime}=5.29\times 10^{-5}J$
(c) We can find the required work done as follows:
$W=\Delta U$
$\implies W=U^{\prime}-U$
We plug in the known values to obtain:
$W=5.29\times 10^{-5}J-2.64\times 10^{-5}J$
$W=2.65\times 10^{-5}J$
