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Question

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

Vedclass · Accepted Answer

$-3 \hat{i} + \hat{j}$

Vedclass · Answer

(B) The electric field $\vec{E}$ is related to the electric 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 $V = \frac{3x^2}{2} - \frac{y^2}{4}$.Calculating the partial derivatives:$E_x = -\frac{\partial V}{\partial x} = -\frac{\partial}{\partial x} \left( \frac{3x^2}{2} - \frac{y^2}{4} ight) = -3x$.$E_y = -\frac{\partial V}{\partial y} = -\frac{\partial}{\partial y} \left( \frac{3x^2}{2} - \frac{y^2}{4} ight) = -\left( -\frac{2y}{4} ight) = \frac{y}{2}$.At the point $(1 \, m, 2 \, m)$,substitute $x = 1$ and $y = 2$:$E_x = -3(1) = -3 \, N/C$.$E_y = \frac{2}{2} = 1 \, N/C$.Thus,the electric field vector is $\vec{E} = -3 \hat{i} + \hat{j} \, N/C$.

Electric potential in a region is varying according to the relation $V = \frac{3x^2}{2} - \frac{y^2}{4}$,where $x$ and $y$ are in $m$ and $V$ is in $V$. Electric field intensity (in $N/C$) at a point $(1 \, m, 2 \, m)$ is:

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