Directions (Q. Nos. 54 to 55) are based on the following paragraph.
Consider a block of conducting m

Q: Why does the current spread hemispherically and not spherically?

Because the points A and D are on the flat surface of the conducting block. The current cannot flow into the air above the block, so it is confined to the hemispherical volume within the conducting material.

Q: Why is the potential due to D negative?

Point D acts as a sink where current is leaving the material. You can think of it as injecting a negative current ($-I$) into the block. Therefore, the electric field points towards D, and the potential is negative.

Q: The book's answer is (c). Why did we get (a)?

The book's solution contains a classic error. It calculates the potential difference $V_B - V_C$ only due to the current entering at A, completely ignoring the effect of the current leaving at D. When you correctly superpose both effects, the potential difference doubles, giving option (a).

Q: How do we know the potential is zero at infinity?

It's a standard convention in electrostatics and steady currents. For a point source spreading out in 3D (or a hemisphere), the electric field falls off as $1/r^2$, so the potential falls off as $1/r$. As $r \to \infty$, the potential naturally approaches zero.

Q: Can we use the formula for resistance $R = \rho L / A$ here?

No, because the cross-sectional area $A$ is not constant. The current spreads out radially, so the area increases as $2\pi r^2$. You must use the microscopic form of Ohm's law ($E = \rho J$) and integrate.

Question

Passage:
Directions (Q. Nos. 54 to 55) are based on the following 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
(a) Take current $I$ entering from $A$ and assume it to spread over a hemispherical surface on the block.
(b) 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 $r$.
(c) From the $r$ dependence of $E(r)$, obtain the potential $V(r)$ at $r$.
(d) Repeat steps (i), (ii) and (iii) for current $I$ leaving $D$ and superpose results for $A$ and $D$.

Question 1:
$\Delta V$ measured between $B$ and $C$ is
(A) $\frac{\rho I}{\pi a} - \frac{\rho I}{\pi(a+b)}$
(B) $\frac{\rho I}{a} - \frac{\rho I}{(a+b)}$
(C) $\frac{\rho I}{2\pi a} - \frac{\rho I}{2\pi(a+b)}$
(D) $\frac{\rho I}{2\pi(a-b)}$

Sigma Percentile · Accepted Answer

### The Setup: Current in a Conducting Block
Imagine a vast conducting block. Current $I$ is injected at point $A$ and extracted at point $D$. We need to find the potential difference between two intermediate points, $B$ and $C$. The problem graciously provides a roadmap: use the **superposition principle**.

### The Master Equation: Potential of a Point Source
When current $I$ enters the block at $A$, it doesn't just flow in a straight line; it spreads out radially in all directions into the bulk of the material. Since $A$ is on the surface, the current spreads over a **hemispherical shell**.

At a distance $r$ from $A$, the surface area of this hemisphere is $2\pi r^2$. Therefore, the current density $J$ (current per unit area) is:
$$J = \frac{I}{2\pi r^2}$$

According to the microscopic form of Ohm's Law, the electric field $E$ is proportional to the current density, with the resistivity $ho$ being the proportionality constant:
$$E = ho J = \frac{ho I}{2\pi r^2}$$

The electric potential $V(r)$ is the negative integral of the electric field. Integrating $E$ from infinity (where potential is zero) to $r$ gives:
$$V(r) = \int_{r}^{\infty} E \, dr = \frac{ho I}{2\pi r}$$

### Applying Superposition
Now, let's find the potential at points $B$ and $C$. We have two sources of current:
1. Current $+I$ entering at $A$.
2. Current $-I$ leaving at $D$ (acting as a sink).

**Potential due to $A$:**
Point $B$ is at a distance $a$ from $A$, and point $C$ is at a distance $a+b$ from $A$.
$$V_B^{(A)} = \frac{ho I}{2\pi a}$$
$$V_C^{(A)} = \frac{ho I}{2\pi (a+b)}$$

**Potential due to $D$:**
Point $C$ is at a distance $a$ from $D$, and point $B$ is at a distance $a+b$ from $D$. Since current is leaving, we use $-I$.
$$V_C^{(D)} = -\frac{ho I}{2\pi a}$$
$$V_B^{(D)} = -\frac{ho I}{2\pi (a+b)}$$

### Final Calculation: The True Potential Difference
By the superposition principle, the total potential at any point is the sum of the potentials due to $A$ and $D$.
$$V_B = V_B^{(A)} + V_B^{(D)} = \frac{ho I}{2\pi a} - \frac{ho I}{2\pi (a+b)}$$
$$V_C = V_C^{(A)} + V_C^{(D)} = \frac{ho I}{2\pi (a+b)} - \frac{ho I}{2\pi a}$$

Notice the beautiful symmetry! $V_C$ is exactly the negative of $V_B$. 
Now, we find the potential difference $\Delta V = V_B - V_C$:
$$\Delta V = \left[ \frac{ho I}{2\pi a} - \frac{ho I}{2\pi (a+b)} ight] - \left[ \frac{ho I}{2\pi (a+b)} - \frac{ho I}{2\pi a} ight]$$
$$\Delta V = 2 	imes \left[ \frac{ho I}{2\pi a} - \frac{ho I}{2\pi (a+b)} ight] = \frac{ho I}{\pi a} - \frac{ho I}{\pi (a+b)}$$

**A Crucial Note on the Provided Answer Key:**
The reference solution provided in the text incorrectly calculates only the potential difference due to the source at $A$ and completely ignores the sink at $D$, leading to option (c). However, a rigorous application of the superposition principle (as explicitly instructed by the problem itself!) yields the result in option (a). Always trust the physics!

Animated Solution for Physics - Current Electricity: Comprehension Passage

$Δ V$ measured between $B$ and $C$ is

Select Answer:

Visualized Solution

The Setup and Superposition

Potential due to a Point Source

Potential at $B$ and $C$ due to Source $A$

Potential at $B$ and $C$ due to Sink $D$

Superposition of Potentials

Calculating Potential Difference

Final Answer

The Sigma Insight: Ohm's Law, Resistance and Electrical Power

The Setup

The Master Equation

Applying Superposition Now, let's find the potential at points $B$ and $C$

Final Calculation

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