Paragraph: 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 '$$\Delta V$$' 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 $$E = \rho j$$, 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 '$$D$$'.
Question: $$\Delta V$$ measured between $$B$$ and $$C$$ is

When a point current source injects a steady current $$I$$ into a large block of uniform resistivity $$\rho$$, the current spreads radially inside the conductor. Close to the entry point $$A$$ the current lines form a hemisphere of radius $$r$$ (only the half-space inside the block is available). The surface area of this hemisphere is $$2\pi r^{2}$$.

Case 1: Current $$I$$ entering at $$A$$
Current density $$j(r)$$ at distance $$r$$ from $$A$$ is obtained from the definition $$j = I/\text{area}$$:

$$j(r)=\frac{I}{2\pi r^{2}}$$

Ohm’s law for a conductor gives the magnitude of the electric field:
$$E(r)=\rho\,j(r)=\frac{\rho I}{2\pi r^{2}}$$

The field is radial, so the potential difference between two points at radii $$r_1$$ and $$r_2$$ from $$A$$ is the line integral of $$E$$:

$$V(r_2)-V(r_1)=-\int_{r_1}^{r_2}E(r)\,dr =-\int_{r_1}^{r_2}\frac{\rho I}{2\pi r^{2}}\,dr =\frac{\rho I}{2\pi}\left(\frac{1}{r_2}-\frac{1}{r_1}\right)$$

Choosing the reference potential at a very large distance (effectively $$r\rightarrow\infty$$, where the term $$1/r$$ vanishes) the absolute potential at a point a distance $$r$$ from $$A$$ is

$$V_A(r)=\frac{\rho I}{2\pi r}$$

In the block the points $$B$$ and $$C$$ lie on the line joining them to $$A$$ at distances
$$AB = a,\quad AC = a+b$$.
Hence

$$V_A(B)=\frac{\rho I}{2\pi a},\qquad V_A(C)=\frac{\rho I}{2\pi(a+b)}$$

Case 2: Current $$I$$ leaving at $$D$$
Exactly the same reasoning applies near $$D$$, but now the current is directed out of the conductor. The electric field near $$D$$ therefore points towards $$D$$, making its contribution to the scalar potential negative. In the given geometry points $$B$$ and $$C$$ are equidistant from $$D$$ (the figure shows them located symmetrically with respect to $$D$$). Consequently, the potentials produced by the exit current at $$B$$ and $$C$$ are equal and cancel in the difference $$\Delta V$$:

$$V_D(B)=V_D(C)\;\Longrightarrow\; V_D(B)-V_D(C)=0$$

Total potential difference $$\Delta V_{BC}$$
Superposing the two cases:

$$\Delta V_{BC}= \big[V_A(B)+V_D(B)\big]-\big[V_A(C)+V_D(C)\big] = V_A(B)-V_A(C)$$

$$\Delta V_{BC}= \frac{\rho I}{2\pi}\left(\frac{1}{a}-\frac{1}{a+b}\right)$$

This matches Option C.

Answer: Option C which is: $$\displaystyle \frac{\rho I}{2\pi a}-\frac{\rho I}{2\pi(a+b)}$$