The potential V is varying with x and y as V | vedclass

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Question

(C) The electric field $\vec{E}$ is related to the potential $V$ by the relation $\vec{E} = -
abla V = -\left( \frac{\partial V}{\partial x} \hat{i}

Vedclass · Accepted Answer

$2\hat{i} - \hat{j} 	ext{ V/m}$

Vedclass · Answer

(C) The electric field $\vec{E}$ is related to the potential $V$ by the relation $\vec{E} = -
abla V = -\left( \frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} ight)$.Given $V = \frac{1}{2}(y^2 - 4x)$.Calculating the partial derivatives:$E_x = -\frac{\partial V}{\partial x} = -\frac{\partial}{\partial x} \left( \frac{1}{2}y^2 - 2x ight) = -(-2) = 2 	ext{ V/m}$.$E_y = -\frac{\partial V}{\partial y} = -\frac{\partial}{\partial y} \left( \frac{1}{2}y^2 - 2x ight) = -\left( \frac{1}{2} \cdot 2y ight) = -y 	ext{ V/m}$.At the point $(1 	ext{ m}, 1 	ext{ m})$:$E_x = 2 	ext{ V/m}$.$E_y = -1 	ext{ V/m}$.Therefore,the electric field is $\vec{E} = 2\hat{i} - \hat{j} 	ext{ V/m}$.

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

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