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Q.

The electric potential V is given as a function of distance x (meter) by $V = (5 x^{2} + 10 x - 9)$ volt. Value of electric field at x = 1 is

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a

$- 20 V / m$

b

$6 V / m$

c

$11 V / m$

d

$- 23 V / m$

answer is A.

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Detailed Solution

Electric field,

$\begin{array}{l} E = - \frac{dV}{dx} = - \frac{d}{dx} (5 x^{2} + 10 x - 9) = - (10 x + 10) \\ ∴ E (x = 1) = - 10 (1 + 1) = - 20 V / m \end{array}$

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