Potential in the x-y plane is given as V = 5( | vedclass

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Question

(B) 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

$-5 \hat{j} \, V/m$

Vedclass · Answer

(B) 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 $V = 5x^2 + 5xy$.Calculating the partial derivatives:$E_x = -\frac{\partial V}{\partial x} = -(10x + 5y)$.At point $(1, -2)$,$E_x = -(10(1) + 5(-2)) = -(10 - 10) = 0$.$E_y = -\frac{\partial V}{\partial y} = -(5x) = -5x$.At point $(1, -2)$,$E_y = -5(1) = -5$.Therefore,the electric field at $(1, -2)$ is $\vec{E} = 0 \hat{i} - 5 \hat{j} = -5 \hat{j} \, V/m$.

Potential in the $x-y$ plane is given as $V = 5(x^2 + xy) \, V$. The electric field at the point $(1, -2)$ will be

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