The electric potential $V$ is given as a function of distance $x$ by $V = 5 x^{2} + 10 x - 9$ . The value of electric field at $x = 1$ is
(All quantities are in SI units.)

- 20

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Solution

The correct option is A $- 20$
$V = 5 x^{2} + 10 x - 9$

$\to E = - \frac{\partial V}{\partial x}^i - \frac{\partial V}{\partial y}^j - \frac{\partial V}{\partial z}^k$

Now, $\frac{\partial V}{\partial x} = 10 x + 10$

and $\frac{\partial V}{\partial y} = \frac{\partial V}{\partial z} = 0$

Hence, $\to E = - (10 x + 10)^i$

For $x = 1$ , $\to E = - 20^i$

Answer-(A)

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