Question

show that the electric field at any point is equal to negative potential gradient at the point​

Answer 1

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Answer 2

General expression for Electric Field Potential is as follows:

$V = \dfrac{Q}{4\pi \epsilon_{0}r}$

Now, differentiation with respect to displacement (i.e. r ).

$\dfrac{dV}{dr} = \dfrac{ d \bigg(\dfrac{Q}{4\pi \epsilon_{0}r} \bigg)}{dr}$

$\implies \dfrac{dV}{dr} = - \dfrac{Q}{4\pi \epsilon_{0} {r }^{2} }$

$\implies \dfrac{dV}{dr} = -E$

$\implies E = - \dfrac{dV}{dr}$

[Hence Proved]