Explanation
Step 1: Uniform Charge distribution on outer surface [Refer Figure]
qQ4πϵ0(1a+1b)
qQ4πϵ0(1a−1b)
qQ4πϵ0(1a2−1b2)
qQ4πϵ0(1a2+1b2)
We know that work done dW=qdV
here , WB→A=qVA−VB=q[Q4πϵ0(OA)−Q4πϵ0(OB)]=qQ4πϵ0[1a−1b] where OA=a,OB=b
From the expression of energy stored in capacitor,
U=12q2C
∴[C]=[q2U=][A2T2ML2T−2]
=[M−1L−2T4A2]
Work done by field W=−ΔU=UA−UB=(−e)VA−(−e)VB
Please disable the adBlock and continue. Thank you.