The most appropriate relationship between electric field and electric potential is given by

E = - \nabla V_{E}

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V_{E} = - \nabla E

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E = \nabla V_{E}

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V = - \nabla E

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Solution

The correct option is A $E = - \nabla V_{E}$
$\to E = - \frac{\partial V}{d x}^i - \frac{\partial V}{d y}^j - \frac{\partial V}{d z}^k$

And, we know that $\nabla = \frac{\partial}{d x} + \frac{\partial}{d y} + \frac{\partial}{d z}$

Hence, we get $\to E = - \nabla V$

Answer-(A)

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Similar questions

For a spontaneous reaction $△$ G, equilibrium constant (K) and $E_{c e l l}^{\circ}$ will be respectively:

A + e^{-} \to A^{-}; E^{0} = - v e

B + e^{-} \to B^{-}; E^{0} = + v e

A + e^{-} \to A^{-}; E^{0} = - v e

B + e^{-} \to B^{-}; E^{0} = + v e

M - 1

M - 2

f_{1}

f_{2}

E_{F e^{2 +} / F e}^{\circ} = - 0.44 V and E_{F e^{3 +} / F e^{2 +}}^{\circ} = 0.77 V

E_{F e^{3 +} / F e}^{\circ}

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