The electric potential as a function of x, y | vedclass

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Question

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

Vedclass · Accepted Answer

$(-20\hat{i} + 30\hat{j})$

Vedclass · Answer

(A) The electric field $\vec{E}$ is related to the electric potential $V$ by the relation $\vec{E} = -
abla V = -(\frac{\partial V}{\partial x}\hat{i} + \frac{\partial V}{\partial y}\hat{j})$.Given $V = 5x^2 - 5y^2$.Calculating the partial derivatives:$\frac{\partial V}{\partial x} = \frac{\partial}{\partial x}(5x^2 - 5y^2) = 10x$.$\frac{\partial V}{\partial y} = \frac{\partial}{\partial y}(5x^2 - 5y^2) = -10y$.Substituting these into the expression for $\vec{E}$:$\vec{E} = -(10x\hat{i} - 10y\hat{j}) = -10x\hat{i} + 10y\hat{j}$.At the point $(2, 3) 	ext{ m}$, substituting $x = 2$ and $y = 3$:$\vec{E} = -10(2)\hat{i} + 10(3)\hat{j} = -20\hat{i} + 30\hat{j} 	ext{ V/m}$.

The electric potential as a function of $x, y$ is given by $V = 5(x^2 - y^2) \text{ V}$. The electric field at a point $(2, 3) \text{ m}$ is . . . . . . $\text{V/m}$.

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