Q144 / 160 · TS EAMCET 2023 (12 May Shift 1) · Physics
The electric potential at a place is varying as $V=\frac{1}{2}\left(y^2-4 x\right)$ volt. Then the electric field at $\mathrm{x}=1 \mathrm{~m}$ and $\mathrm{y}=1 \mathrm{~m}$ is
- A$2 \hat{i}+\hat{j} V m^{-1}$
- B$-2 \hat{i}+\hat{j} V m^{-1}$
- Correct answerC$2 \hat{i}-\hat{j} V m^{-1}$
- D$-2 \hat{i}+2\hat{j} V m^{-1}$
Solution
Electric potential is given by
$\begin{aligned}
& V=\frac{1}{2}\left(y^2-4 x\right) \\
& E_x=\frac{d V}{d x}=-\left[\frac{1}{2} x-4\right]=2 \mathrm{~V} / \mathrm{m} \\
& E_y=\frac{-d V}{d y}=\frac{1}{2} \times 2 y=y=-1 V / m
\end{aligned}$
The electric field at $x=1 \mathrm{~m}$ and $y=1 \mathrm{~m}$ is
$\begin{aligned}
& E=E_x \hat{i}+E_y \hat{j} \\
& =2 \hat{i}-\hat{j}
\end{aligned}$
Hence, the correct answer is option: 'C'.