The electric potential at a point is given by $V = - 6 x^{2}$ The electric field strength at x=2 is

-24 Vm^-1

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12 Vm^-1

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24 Vm^-1

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4 Vm^-1

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Solution

The correct option is C
24 Vm^-1

$E = - \frac{d v}{d x} = - \frac{d}{d x} (- 6 x^{2}) = 12 x E | x = 2 = 12 (2) = 24 V m^{- 1}$

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