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A surface charge density of 2.10×10-9 C/cm2 exists on a circle-shaped disc. If the disc's radius increases by a factor of 2 (but the total charge stays the same), what is the surface charge density?

Question

A surface charge density of 2.10×10-9 C/cm2 exists on a circle-shaped disc. If the disc's radius increases by a factor of 2 (but the total charge stays the same), what is the surface charge density?

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Solution

The surface charge density (σ) is defined as the total charge (Q) divided by the area (A) over which the charge is spread. In mathematical terms, this is expressed as:

σ = Q / A

Given that the total charge remains the same, but the radius of the disc doubles, the new area of the disc will be four times the original area (since the area of a circle is proportional to the square of the radius, A = πr²).

Therefore, the new surface charge density (σ') will be one fourth of the original surface charge density.

So, σ' = σ / 4

Given that the original surface charge density (σ) is 2.10×10^-9 C/cm², the new surface charge density (σ') will be:

σ' = 2.10×10^-9 C/cm² / 4 = 0.525×10^-9 C/cm².

This problem has been solved

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