The electric potential at a place is varying | vedclass

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Question

(C) 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

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

Vedclass · Answer

(C) 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 $V = \frac{1}{2}y^2 - 2x$.Calculating the partial derivatives:$\frac{\partial V}{\partial x} = \frac{\partial}{\partial x} (\frac{1}{2}y^2 - 2x) = -2$.$\frac{\partial V}{\partial y} = \frac{\partial}{\partial y} (\frac{1}{2}y^2 - 2x) = y$.Thus,$\vec{E} = -(-2 \hat{i} + y \hat{j}) = 2 \hat{i} - y \hat{j}$.At the point $(x = 1 	ext{ m}, y = 1 	ext{ m})$:$\vec{E} = 2 \hat{i} - (1) \hat{j} = 2 \hat{i} - \hat{j} 	ext{ V m}^{-1}$.

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

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