Q178 / 200 · MHT CET - Medical 2026 (26 April Shift 1) · Physics
The electric potential 'V' as a function of distance x (in metre) is $V = 4x^2 + 6x - 8$ volt. Electric field at $x = 2$ m is
- A$-11\ \text{V/m}$
- Correct answerB$-22\ \text{V/m}$
- C$-16\ \text{Vm}$
- D$-8\ \text{V/m}$
Solution
The relation between electric field $E$ and electric potential $V$ is given by $E = -\dfrac{dV}{dx}$.
Given $V = 4x^2 + 6x - 8$.
Differentiating with respect to $x$:
$\dfrac{dV}{dx} = \dfrac{d}{dx}(4x^2 + 6x - 8) = 8x + 6$
Therefore, $E = -(8x + 6)$.
At $x = 2$ m:
$E = -(8(2) + 6) = -(16 + 6) = -22\ \text{V/m}$.
Answer: $-22\ \text{V/m}$
Given $V = 4x^2 + 6x - 8$.
Differentiating with respect to $x$:
$\dfrac{dV}{dx} = \dfrac{d}{dx}(4x^2 + 6x - 8) = 8x + 6$
Therefore, $E = -(8x + 6)$.
At $x = 2$ m:
$E = -(8(2) + 6) = -(16 + 6) = -22\ \text{V/m}$.
Answer: $-22\ \text{V/m}$
Hence, the correct answer is option: 'B'.