Q176 / 180 · KCET 2009 · Physics

The electric potential at any point $x, y, z$ in metres is given by $\mathrm{V}=3 \mathrm{x}^{2}$. The electric field at a point $(2,0,1)$ is

  • A
    $12 \mathrm{Vm}^{-1}$
  • B
    $-6 \mathrm{Vm}^{-1}$
  • C
    $6 \mathrm{Vm}^{-1}$
  • Correct answer
    D
    $-12 \mathrm{Vm}^{-1}$

Solution

Here, electric potential $$ V=3 x^{2} $$ Electric field, $$ \begin{aligned} \mathrm{E}=-\frac{\partial \mathrm{V}}{\partial \mathrm{x}} &=-\frac{\partial}{\partial \mathrm{x}}\left(3 \mathrm{x}^{2}\right)=-6 \mathrm{x} \\ \therefore \quad & \mathrm{E}_{(2,0,1)} =-12 \mathrm{Vm}^{-1} \end{aligned} $$

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

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