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NEET 2011
Exam
NEET
Year
2011
Subject
Physics
Chapter
Electrostatics

NEET 2011 · Physics · Electrostatics

The electric potential V at any point (x, y, z) , all in metres in...

The electric potential VV at any point (x,y,z)(x, y, z), all in metres in space is given by V=4x2V=4 x^2 volt. The electric field at the point (1,0,2)(1,0,2) in volt/metre is

Options

  1. A8 along positive XX-axis
  2. B16 along negative XX-axis
  3. C16 along positive XX-axis
  4. D8 along negative XX-axis

Correct answer

D. 8 along negative XX-axis

Step-by-step solution

We know that E=−\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]SoSoE=−i^∂V∂x=−i^∂∂x(4x2)=−8xi^Vm−1E(1,0,2)=−8i^Vm−1\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}$
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