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JIPMER 2019
Exam
JIPMER
Year
2019
Subject
Physics
Chapter
Electromagnetic Waves

JIPMER 2019 · Physics · Electromagnetic Waves

Potential difference is given as V(x)=-x^2 y volt. Find electric...

Potential difference is given as V(x)=−x2yV(x)=-x^2 y volt. Find electric field at a point (1,2)(1,2) ?

Options

  1. Ai^+4j^V/m\hat{\mathrm{i}}+4 \hat{\mathrm{j}} \mathrm{V} / \mathrm{m}
  2. B−4i^−j^V/m-4 \hat{\mathbf{i}}-\hat{\mathbf{j}} \mathrm{V} / \mathrm{m}
  3. C4i^+j^V/m4 \hat{\mathbf{i}}+\hat{\mathbf{j}} \mathrm{V} / \mathrm{m}
  4. D4i^−j^V/m4 \hat{\mathbf{i}}-\hat{\mathbf{j}} \mathrm{V} / \mathrm{m}

Correct answer

C. 4i^+j^V/m4 \hat{\mathbf{i}}+\hat{\mathbf{j}} \mathrm{V} / \mathrm{m}

Step-by-step solution

Potential difference V(x)=−x2yVE=−ΔV=−[i^∂∂x(−x2y)+j^∂∂y(−x2y)]\begin{aligned} V(x) & =-x^2 y V \\ E & =-\Delta V=-\left[\hat{i} \frac{\partial}{\partial x}\left(-x^2 y\right)+\hat{j} \frac{\partial}{\partial y}\left(-x^2 y\right)\right] \end{aligned}=−[−2xyi^−x2j^]=2xyi^+x2j^ E at (1,2)=2×1×2i^+12j^=4i^+j^V/m\begin{aligned} & =-\left[-2 x y \hat{i}-x^2 \hat{j}\right]=2 x y \hat{i}+x^2 \hat{j} \\ \text{ E at }(1,2) & =2 \times 1 \times 2 \hat{i}+1^2 \hat{j} \\ & =4 \hat{i}+\hat{j} V / m\end{aligned}$
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