1. Concept: For a point charge, the electric field (\( E \)) decays by an inverse-square law, while the electric potential (\( V \)) decays inversely with linear distance.
2. Formula: $$ E \propto \frac{1}{r^2} \quad \text{and} \quad V \propto \frac{1}{r} $$
3. Solution: - If the new field is \( E/4 \), this means the distance must have been doubled (since \( 1/(2r)^2 = 1/4r^2 \)). Let the new distance be \( r' = 2r \).
- Since potential depends on \( 1/r \), applying the new doubled distance gives \( V' = \frac{k q}{2r} = \frac{1}{2} \left( \frac{k q}{r} \right) \).
- Therefore, the new potential is \( V/2 \).
4. Why other options are incorrect: Option A falsely assumes V decreases at the identical squared rate as E. Option D incorrectly implies that moving further away increases the potential.
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