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

The electric field \(\vec{E}\) is given by the negative gradient of the electric potential \(V\):

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

Given \(V = 5(x^2 - y^2)\), we find the partial derivatives:

\(\dfrac{\partial V}{\partial x} = 10x\)

\(\dfrac{\partial V}{\partial y} = -10y\)

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 \((2, 3)\), substitute \(x = 2\) and \(y = 3\):

\(\vec{E} = -10(2) \hat{i} + 10(3) \hat{j} = -20\hat{i} + 30\hat{j}\) V/m

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