 Solution

ID: past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2015-resit/1/solution

r > 0

. For a positive point charge

q

V (r) = \frac{q}{4 π ϵ _{0} r}, E (r) = \frac{q}{4 π ϵ _{0} r ^{2}} \hat{r} .

The field is radial and outward. Since a radial scalar function has gradient

\nabla V = (d V / d r) \hat{r}

- \nabla V = - (- \frac{q}{4 π ϵ _{0} r ^{2}}) \hat{r} = E .

Thus the two expressions have the required vector relationship. Adding an arbitrary constant to

V

would leave the electric field unchanged; the chosen reference at infinity fixes that constant. The point-charge singularity itself is excluded from the ordinary derivative calculation.

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