Consider a block of conducting material of resistivity rho shown in the figure. Current I enters at

Q: Why do we assume a hemispherical surface?

Because the current enters at a point on a flat surface and spreads out uniformly into the bulk of the material. Since it cannot flow upwards into the air, it forms half-spheres (hemispheres) expanding downwards and outwards.

Q: What is the microscopic form of Ohm's law?

It is $E = \rho J$. This form relates the electric field at a specific point to the current density at that exact same point, unlike $V = IR$ which applies to a whole macroscopic component.

Q: Why isn't the area $4\pi r^2$?

The current is constrained to flow only within the conducting block (below the surface). It cannot flow into the non-conducting air above, so it only forms a hemisphere, not a full sphere.

Q: Does the electric field depend on the shape of the block?

As long as the block is large enough compared to the distance $r$ that the boundaries don't restrict or distort the flow, the hemispherical assumption holds perfectly true.

Q: How will this help find the potential difference?

Potential difference is the negative line integral of the electric field ($V = -\int E \cdot dr$). By knowing $E(r)$, we can integrate it along the path between any two points to find the voltage drop.

Question

Passage:
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:
For current entering at $A$, the electric field at a distance $r$ from $A$ is
(A) $\frac{\rho I}{8\pi r^2}$
(B) $\frac{\rho I}{r^2}$
(C) $\frac{\rho I}{2\pi r^2}$
(D) $\frac{\rho I}{4\pi r^2}$

Sigma Percentile · Accepted Answer

### The Spreading Current: Finding the Electric Field in a Conductor

Imagine you are observing a large block of conducting material. A steady current $I$ is injected into this block at a specific point, let's call it $A$. What happens to the current once it enters? It doesn't just travel in a straight, narrow line. Because the block is a large, continuous medium, the current spreads out uniformly in all available directions.

Since the current enters at a point on the flat top surface of the block, it can only spread downwards and outwards into the bulk of the material. This uniform spreading creates **hemispherical shells** of current flow expanding away from point $A$.

### Current Density: The Missing Link

To find the electric field at a certain distance $r$ from the entry point, we first need to understand how densely the current is packed at that location. This concept is known as **current density**, denoted by the vector $\vec{J}$.

Mathematically, the magnitude of current density is simply the total current $I$ divided by the cross-sectional area $A$ it passes through:
$$J = \frac{I}{A}$$

At a distance $r$ from point $A$, the current has spread over a hemispherical surface. The surface area of a full sphere is $4\pi r^2$, so the area of our hemisphere is exactly half of that:
$$A = 2\pi r^2$$

Substituting this area into our current density formula, we get:
$$J = \frac{I}{2\pi r^2}$$

### Ohm's Law in Microscopic Form

You are likely familiar with the macroscopic form of Ohm's law, $V = IR$. However, to find the electric field at a specific point, we must use the **microscopic form of Ohm's law**. This fundamental relation connects the electric field $\vec{E}$ at a point to the current density $\vec{J}$ at that same point, using the material's resistivity $\rho$:
$$E = \rho J$$

### The Final Expression

Now, the final step is a simple substitution. We take our expression for the current density $J$ and plug it into the microscopic Ohm's law:
$$E = \rho \left( \frac{I}{2\pi r^2} \right)$$

Rearranging this slightly, we arrive at our final answer:
$$E = \frac{\rho I}{2\pi r^2}$$

This elegant equation reveals a crucial physical insight: the electric field inside the conductor decreases inversely with the square of the distance from the point where the current enters. This result is the stepping stone for calculating the potential difference between any two points on the block's surface by integrating the electric field along the path between them.

Animated Solution for Physics - Current Electricity: Comprehension Passage

For current entering at $A$ , the electric field at a distance $r$ from $A$ is

Select Answer:

Visualized Solution

Current Spreading

Current Density

Hemispherical Area

Ohm's Law (Microscopic)

Electric Field

Conclusion

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

The Spreading Current

Current Density

Ohm's Law in Microscopic Form

The Final Expression

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Animated Solution for Physics - Current Electricity: Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For current entering at A, the electric field at a distance r from A is

Select Answer:

Visualized Solution

Current Spreading

Current Density

Hemispherical Area

Ohm's Law (Microscopic)

Electric Field

Conclusion

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

The Spreading Current

Current Density

Ohm's Law in Microscopic Form

The Final Expression

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔV measured between B and C is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Space between two concentric conducting spheres of radii a and b(b>a) is filled with a medium of resistivity ρ. The resistance between the two spheres (in ohm) will be

For current entering at $A$ , the electric field at a distance $r$ from $A$ is

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Space between two concentric conducting spheres of radii $a$ and $b (b > a)$ is filled with a medium of resistivity $ρ$ . The resistance between the two spheres (in ohm) will be