The electric potential ' V ' is given as a function of distance ' x ' (metre) by V = (4x² + 8x - 3)V. The value of electric field at x = 0.5 m, in V/m is (a) -16 (b) -12 (c) 0 (d) +12

Answer: (b). E = -dV/dx = -(8x + 8) At x = 0.5: E = -(4 + 8) = -12 V/m So the answer is (b) -12.

The electric potential ' V ' is given as a function of distance ' x '… | MHT CET 2025 (19 Apr, Shift 2)

Q32MHT CET 2025 · 19 APR · SHIFT 2Easy

Electric Potential & Equipotentials

The electric potential ' V ' is given as a function of distance ' $x$ ' (metre) by $V = \left(4x^2 + 8x - 3\right)V$ . The value of electric field at $x = 0.5\text{ m}$ , in $\text{V/m}$ is

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Answer: (b)

$E = -\frac{dV}{dx} = -(8x + 8)$
At $x = 0.5$ : $E = -(4 + 8) = -12\text{ V/m}$
So the answer is (b) $-12$ .

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