The potential is varying with distance (x, y) | vedclass

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(C) The electric field $\vec{E}$ is given by the negative gradient of the potential $V$: $\vec{E} = -\vec{
abla} V = -\left( \frac{\partial V}{\parti

Vedclass · Accepted Answer

$2 \hat{i} - \hat{j} 	ext{ Vm}^{-1}$

Vedclass · Answer

(C) The electric field $\vec{E}$ is given by the negative gradient of the potential $V$: $\vec{E} = -\vec{
abla} V = -\left( \frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} ight)$.Given $V = \frac{1}{2} y^2 - 2x$.Calculating partial derivatives:$\frac{\partial V}{\partial x} = -2$$\frac{\partial V}{\partial y} = y$Substituting these into the electric field formula:$\vec{E} = -(-2 \hat{i} + y \hat{j}) = 2 \hat{i} - y \hat{j}$.At the point $(x = 1 	ext{ m}, y = 1 	ext{ m})$:$\vec{E} = 2 \hat{i} - (1) \hat{j} = 2 \hat{i} - \hat{j} 	ext{ Vm}^{-1}$.

The potential is varying with distance $(x, y)$ as $V = \frac{1}{2} (y^2 - 4x) \text{ V}$. The electric field at $x = 1 \text{ m}$ and $y = 1 \text{ m}$ is:

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