The potential V varies with x and y as V = fr | vedclass

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

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 $\vec{E} = -
abla V = -(\frac{\partial V}{\partial x}\hat{i} + \frac{\partial V}{\partial y}\hat{j})$.Given $V = \frac{1}{2}(y^2 - 4x)$.Calculating the partial derivative with respect to $x$:$E_x = -\frac{\partial V}{\partial x} = -\frac{\partial}{\partial x} [\frac{1}{2}(y^2 - 4x)] = -\frac{1}{2}(-4) = 2 	ext{ V/m}$.Calculating the partial derivative with respect to $y$:$E_y = -\frac{\partial V}{\partial y} = -\frac{\partial}{\partial y} [\frac{1}{2}(y^2 - 4x)] = -\frac{1}{2}(2y) = -y 	ext{ V/m}$.At the point $(1 	ext{ m}, 1 	ext{ m})$,we have $x = 1$ and $y = 1$.Substituting $y = 1$ into the expression for $E_y$:$E_y = -1 	ext{ V/m}$.Thus,the electric field vector is $\vec{E} = E_x\hat{i} + E_y\hat{j} = 2\hat{i} - 1\hat{j} 	ext{ V/m}$.

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

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