Consider a block of conducting material of resistivity `'rho'` shown in the figure. Current `'I'` enters at 'A' and leaves from 'D'. We apply superposition principle to find voltage `DeltaV` developed between 'B' and 'C'. The calculation is done in the following steps:
(i) Take current 'I' entering from 'A' and assume it to spread over a hemispherical surface in the block.
(ii) Calculate field E(r) at distance 'r' from A by using Ohm's law `E = rhoj,` where j is the current per unit area at 'r'.
(iii) From the 'r' dependence of E(r), obtain the potential V(r) at r.
(iv) Repeat (i), (ii) and (iii) for current 'I' leaving 'D' and superpose results for 'A' and 'D'.

`Delta V measured between B and C is

Options

A. `(pI)/(2pi (a-b))`
B. `(pI)/(pi a)-(p i)/(pi (a+b))`
C. `(P I)/a-(P I)/((a+b))`
D. `(pI)/(2pia)-(P I)/(2 pi (a+b))`

Explanation

Current entering A is distributed over the hemispherical surface of area `2pir^2`
Hence current density `j=I/(2pir^2)`
Resistance `=(pl)/(area)=(pr)/(2pir^2)` , Electric field `E=pj=(pI)/(2pir^2)`
Hence potential difference,
`V_B-V_C=Delta V= int_(a+b)^a -Edr implies Delta V=(-D)/(2pi) int_(a+b)^a 1/r^2 dr=(-D)/(2pir^2) [-1/r]_(a+b)^a , Delta V=(Ip)/(2pi)[1/a-1/(a+b)]`
For current leaving D, we get same `Delta V` between B and C. Superposing both results, `Delta V=(Ip)/(2pi)[1/a-1/(a+b)]`