Q74 / 120 · NEET 2011 (Mains) · Physics

The electric potential $V$ at any point $(x, y, z)$, all in metres in space is given by $V=4 x^2$ volt. The electric field at the point $(1,0,2)$ in volt/metre is

  • A
    8 along positive $X$-axis
  • B
    16 along negative $X$-axis
  • C
    16 along positive $X$-axis
  • Correct answer
    D
    8 along negative $X$-axis

Solution

We know that $\mathrm{E}=-\left[\hat{i} \frac{\partial V}{\partial x}+\hat{j} \frac{\partial V}{\partial y}+\mathrm{k} \frac{\partial V}{\partial z}\right]$ So $\begin{aligned} & \mathrm{E}=-\hat{i} \frac{\partial V}{\partial x}=-\hat{i} \frac{\partial}{\partial x}\left(4 x^2\right) \\ &=-8 x \hat{i} \mathrm{Vm}^{-1} \\ & \mathrm{E}_{(1,0,2)}=-8 \hat{i} \mathrm{Vm}^{-1} \end{aligned}$

Hence, the correct answer is option: 'D'.

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