A charge of +5µC is placed in free space. P and Q are two points at 3

1.	A charge of +5µC is placed in free space. P and Q are two points at 3 mm and 5 mm respectively from 5µC.1. The work done to bring a + IC charge from infinity to a point ‘P’ is calledcapacitancedielectric constantpotential energyelectric potential2. Calculate the work done in above process.3. Calculate the work done to move a +IC charge from ‘P’to ‘Q’.
Answer» 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

A charge of +5µC is placed in free space. P and Q are two points at 3 mm and 5 mm respectively from 5µC.1. The work done to bring a + IC charge from infinity to a point ‘P’ is calledcapacitancedielectric constantpotential energyelectric potential2. Calculate the work done in above process.3. Calculate the work done to move a +IC charge from ‘P’to ‘Q’.

Answer»

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