Physics Q37 – JEE-MAIN 2026

Home Physics Electric Charges and FieldsQuestion #99

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.

Clue 1 of 3: Relationship between Electric Field and Potential

The electric field is related to the electric potential by the negative gradient of the potential.

🧠 Full Solution Path

Step 1: Recall the relationship between electric field and potential✦ Active

The electric field $\vec{E}$ is the negative gradient of the electric potential $V$ . For a potential $V (x, y)$ , the electric field in Cartesian coordinates is given by:

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

Step 2: Calculate the partial derivatives of the potential○ Expand

Given $V = 5 (x^{2} - y^{2}) = 5 x^{2} - 5 y^{2}$ . Calculate the partial derivative with respect to $x$ :

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

Calculate the partial derivative with respect to $y$ :

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

💡 Teacher's Secret Hint

Remember to treat other variables as constants during partial differentiation.

Step 3: Substitute derivatives and evaluate at the given point○ Expand

Substitute the partial derivatives into the electric field equation:

\vec{E} = - (10 x \hat{i} - 10 y \hat{j}) = - 10 x \hat{i} + 10 y \hat{j}

Evaluate $\vec{E}$ at the point $(x, y) = (2, 3)$ :

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

This result matches option 1.

💡 Teacher's Secret Hint

Pay close attention to the negative sign in the gradient formula.

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Physics Question 37 – JEE-MAIN 2026

Clue 1 of 3: Relationship between Electric Field and Potential

🧠 Full Solution Path