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}JEE Main 2026 (Online) 6th April Evening Shift – Physics PYQ: Electric Potential & Field
\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}
This explanation is clear and effective because it doesn’t just give the correct answer — it walks through the reasoning step by step, from the relation \(\vec{E} = -\nabla V\) to evaluating the components at the given point. That makes it much easier for students to understand the concept instead of just memorizing the result. I also appreciate how neatly the final vector form is presented, since that helps connect the math with the physical meaning. Clear educational content like this is always valuable, and it also reminds me how useful good visual tools can be for learning and presentation — whether it’s equations or creative content, platforms like PicWav（[https://picwav.com](https://picwav.com)） show how much clearer ideas can feel when presented in a more engaging way. Overall, a simple but very helpful explanation.