The electric potential as a function of x, y is given by V = 5(x² – y²) V. The electric field at a point (2, 3) m is __ V/m.

\begin{aligned}
&\text{\textbf{Question:}} \\
&\text{The electric potential as a function of } x, y \text{ is given by } V = 5(x^2 - y^2)\text{V}. \\
&\text{The electric field at a point } (2, 3)\text{m} \text{ is \underline{\hspace{1cm}} V/m.} \\
\\
&\text{\textbf{A) }} \left(-20\hat{i} + 30\hat{j}\right) \\
&\text{\textbf{B) }} \left(20\hat{i} - 30\hat{j}\right) \\
&\text{\textbf{C) }} \left(20\hat{i} + 45\hat{j}\right) \\
&\text{\textbf{D) }} \left(-4\hat{i} + 6\hat{j}\right)
\end{aligned}

\begin{aligned}
&\text{\textbf{Explanation}} \\
&\text{The electric field } \vec{E} \text{ is the negative gradient of the electric potential } V: \\
&\vec{E} = -\nabla V = -\left( \frac{\partial V}{\partial x}\hat{i} + \frac{\partial V}{\partial y}\hat{j} + \frac{\partial V}{\partial z}\hat{k} \right) \\
\\
&\text{The electric potential is given as } V = 5\left(x^2 - y^2\right) = 5x^2 - 5y^2. \\
\\
&\text{The } x\text{-component of the electric field is:} \\
&E_x = -\frac{\partial V}{\partial x} = -\frac{\partial}{\partial x}\left(5x^2 - 5y^2\right) = -(10x) = -10x \\
\\
&\text{The } y\text{-component of the electric field is:} \\
&E_y = -\frac{\partial V}{\partial y} = -\frac{\partial}{\partial y}\left(5x^2 - 5y^2\right) = -(-10y) = 10y \\
\\
&\text{\textbf{At Point } } (2, 3): \\
&E_x = -10(2) = -20 \text{ V/m} \\
&\Rightarrow E_y = 10(3) = 30 \text{ V/m} \\
\\
&\text{So, the resulting electric field vector is:} \\
&\vec{E} = -20\hat{i} + 30\hat{j} \\
\\
&\text{Therefore, the correct option is \textbf{(A)}.}
\end{aligned}