AP EAMCET · PHYSICS · Electrostatics
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
- A \(2 \hat{\mathrm{i}}+\hat{\mathrm{j}} \mathrm{Vm}^{-1}\)
- B \(-2 \hat{i}+\hat{j} V m^{-1}\)
- C \(2 \hat{\mathrm{i}}-\hat{\mathrm{j}} \mathrm{Vm}^{-1}\)
- D \(-2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}} \mathrm{Vm}^{-1}\)
Answer & Solution
Correct Answer
(C) \(2 \hat{\mathrm{i}}-\hat{\mathrm{j}} \mathrm{Vm}^{-1}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & V=\frac{1}{2}\left(y^2-4 x\right) \\ & E=-\vec{\nabla} V \\ & =-\left[-\frac{4}{2} \hat{i}+\frac{2 y}{2} \hat{j}\right] \\ & =-[-2 \hat{i}+y \hat{j}] \text { at } x=1, y=1 \\ & =2 \hat{i}-\hat{j} V / m\end{aligned}\)
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