The potential is varying with distance $(\mathrm{x})$ as $\mathrm{V}=\frac{1}{2}$ $\left(\mathrm{y}^2-4…

The potential is varying with distance $(\mathrm{x})$ as $\mathrm{V}=\frac{1}{2}$ $\left(\mathrm{y}^2-4 \mathrm{x}\right)$ volt. The electric field at $\mathrm{x}=1 \mathrm{~m}$ and $\mathrm{y}=1 \mathrm{~m}$ is
  1. $2 \hat{\mathrm{i}}+\hat{\mathrm{j}} \mathrm{Vm}^{-1}$
  2. $-2 \hat{i}+\hat{j} V m^{-1}$
  3. $2 \hat{\mathrm{i}}-\hat{\mathrm{j}} \mathrm{Vm}^{-1}$
  4. $-2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}} \mathrm{Vm}^{-1}$

Solution

$\begin{aligned} & V=\frac{1}{2}\left(y^2-4 x\right) \\ & E=-\vec{\nabla} V \\ & =-\left[-\frac{4}{2} \hat{i}+\frac{2 y}{2} \hat{j}\right] \\ & =-[-2 \hat{i}+y \hat{j}] \text { at } x=1, y=1 \\ & =2 \hat{i}-\hat{j} V / m\end{aligned}$

Asked in: AP EAMCET 2023 (17 May Shift 2)

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