Electric Field from Potential Gradient in Two Dimensions

The relationship between the electric field $\vec{E}$ and the electric potential $V$ is given by the negative gradient of the potential:

$\vec{E} = -\nabla V = -\left( \frac{\partial V}{\partial x}\hat{i} + \frac{\partial V}{\partial y}\hat{j} \right)$

Given the electric potential function: $V(x, y) = 5(x^2 - y^2) = 5x^2 - 5y^2$

First, find the partial derivatives of $V$ with respect to $x$ and $y$: $\frac{\partial V}{\partial x} = \frac{\partial}{\partial x}(5x^2 - 5y^2) = 10x$

$\frac{\partial V}{\partial y} = \frac{\partial}{\partial y}(5x^2 - 5y^2) = -10y$

Now, substitute these derivatives into the electric field formula: $\vec{E} = -(10x \hat{i} - 10y \hat{j}) = -10x \hat{i} + 10y \hat{j}$

Evaluating the electric field at the point $(x, y) = (2, 3)\text{ m}$: $\vec{E}(2, 3) = -10(2)\hat{i} + 10(3)\hat{j} = (-20\hat{i} + 30\hat{j})\text{ V/m}$

Thus, the electric field at the point $(2, 3)\text{ m}$ is $(-20\hat{i} + 30\hat{j})\text{ V/m}$.

Correct Option: A