First slide

Electrostatic potential

Question

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

Easy

A

-20 V/m
B

6 V/m
C

11 V/m
D

-23 V/m

Solution

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

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