1. Electric potential.
2. Work done w = VQ
\(=\frac{1}{4\pi\Sigma_0}\frac{q}{r}\times Q\) = \(\frac{9\times10^9\times5\times10^{-6}\times1}{3\times10^{-3}}\) = 15 J
3. The potential energy at p,
PE1 = \(\frac{9\times10^9\times5\times10^{-6}\times1}{3\times10^{-3}}\) = 15 V
The potential energy at Q
PE2 = CQ = \(\frac{9\times10^9\times5\times10^{-6}\times1}{5\times10^{-3}}\)
PE2 = 9V
Work done to move IC from P to Q, W = PE2 – PE1
= 9 – 15
w = -6 J