If the distance from a point charge is doubled, how does the electric field change relative to its original value?

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Multiple Choice

If the distance from a point charge is doubled, how does the electric field change relative to its original value?

Explanation:
The key idea is that the electric field from a point charge follows the inverse-square law: E ∝ 1/r^2. If you double the distance, you replace r with 2r, giving E' = kq/(2r)^2 = kq/(4r^2) = (1/4) E. So the field becomes one-quarter of its original value when the distance is doubled. This falling-off with distance is a signature of point sources in three-dimensional space.

The key idea is that the electric field from a point charge follows the inverse-square law: E ∝ 1/r^2. If you double the distance, you replace r with 2r, giving E' = kq/(2r)^2 = kq/(4r^2) = (1/4) E. So the field becomes one-quarter of its original value when the distance is doubled. This falling-off with distance is a signature of point sources in three-dimensional space.

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