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. (a) (-20i + 30j) (b) (20i - 30j) (c) (20i + 45j) (d) (-4i + 6j)

Answer: (a). Ex = -(∂ V)/(∂ x) = -10x, Ey = -(∂ V)/(∂ y) = +10y At (2, 3): Ex = -20, Ey = 30 So the answer is (a) (-20 i + 30 j) V/m.

The electric potential as a function of x, y is given by V = 5(x² -… | JEE Main 2026 (6 Apr, Shift 2)

Q5JEE Main 2026 · 6 APR · SHIFT 2Easy

Electric Potential & Equipotentials

The electric potential as a function of $x, y$ is given by $V = 5\left(x^2 - y^2\right)V$ . The electric field at a point $(2, 3)\text{ m}$ is $\underline{\qquad}\ \text{V/m}$ .

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Answer: (a)

$E_x = -\dfrac{\partial V}{\partial x} = -10x$ , $E_y = -\dfrac{\partial V}{\partial y} = +10y$
At $(2, 3)$ : $E_x = -20$ , $E_y = 30$
So the answer is (a) $(-20\hat i + 30\hat j)$ V/m.

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