Consider a thin square sheet of side L and thickness t, made of a material of resistivity rho. The r

Q: Why is the length $L$ and not $t$?

Because the current flows from one shaded face to the opposite shaded face. The distance between these two faces is the side length of the square, which is $L$.

Q: Why is the area $L \times t$?

The current enters through the shaded face. The dimensions of this rectangular face are the side of the square ($L$) and the thickness of the sheet ($t$).

Q: Does this mean a larger square sheet has the same resistance?

Yes! As long as the thickness $t$ and the material ($\rho$) remain constant, the resistance between the opposite edges of a square sheet is always the same, regardless of how large the square is.

Q: What if the sheet was rectangular, say $L \times W$?

If the current flowed along the length $L$, the area would be $W \times t$. The resistance would be $R = \rho \frac{L}{W \times t}$, which would depend on both $L$ and $W$.

Q: What if the current flowed between the top and bottom faces?

In that case, the length of the current path would be the thickness $t$, and the cross-sectional area would be the area of the square, $L^2$. The resistance would then be $R = \rho \frac{t}{L^2}$.

Question

Consider a thin square sheet of side $L$ and thickness $t$, made of a material of resistivity $\rho$. The resistance between two opposite faces, shown by the shaded areas in the figure is

Options:

(A) directly proportional to $L$

(B) directly proportional to $t$

(C) independent of $L$

(D) independent of $t$

Sigma Percentile · Accepted Answer

### Visualizing the Conductor

Imagine you are holding a thin square sheet made of a resistive material. The sheet has a side length of $L$ and a very small thickness $t$. The problem states that current flows between the two opposite shaded faces.

To solve this, we need to understand how the geometry of the sheet affects the flow of electrons. The fundamental property that governs this is electrical resistance.

### The Master Equation for Resistance

The resistance $R$ of any uniform conductor is given by the classic formula:

$$R = ho \frac{l}{A}$$

Here, $ho$ is the resistivity of the material (an intrinsic property), $l$ is the length the current must travel, and $A$ is the cross-sectional area through which the current flows. Our goal is to find $l$ and $A$ for this specific setup.

### Plugging in the Geometry

Let's trace the path of the current. It enters one shaded face and travels straight across the sheet to exit from the opposite shaded face. The distance it covers is exactly the side length of the square sheet. Therefore, the effective length of our conductor is:

$$l = L$$

Next, we look at the face where the current enters. This face is a rectangle. Its height is the thickness of the sheet $t$, and its width is the side of the square $L$. Thus, the cross-sectional area is:

$$A = L 	imes t$$

### The Elegant Cancellation

Now, we substitute our geometric findings back into the master equation:

$$R = ho \frac{L}{L 	imes t}$$

Look closely at this expression. The side length $L$ appears in both the numerator (as the length of the path) and the denominator (as part of the cross-sectional area). They perfectly cancel each other out!

$$R = \frac{ho}{t}$$

### A New Perspective

This is a beautiful and somewhat counterintuitive result. It tells us that the resistance of this square sheet depends **only** on its thickness $t$ and the material's resistivity $ho$. It is completely **independent** of the side length $L$.

Whether the square sheet is $1 	ext{ cm}$ wide or $1 	ext{ km}$ wide, as long as it remains a square with the same thickness, the resistance between opposite faces will be exactly the same. Therefore, the correct answer is that the resistance is independent of $L$.

Animated Solution for Physics - Current Electricity: Consider a thin square sheet of side $L$ and thickness $t$ , made of a material of resistivity $ρ$ . The resistance between two opposite faces, shown by the shaded areas in the figure is

Select Answer:

Visualized Solution

Visualizing the Setup

Formula for Resistance

Determining Length

Determining Area

Substituting Values

Simplifying the Expression

Conclusion

Food for Thought

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

Visualizing the Conductor

The Master Equation for Resistance

Plugging in the Geometry

The Elegant Cancellation

A New Perspective

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