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Physics · Electrostatics

JEE Main 2026 — 6 April, Evening Shift — Question 10

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

  1. Option A:

    (−20i^+30j^)(-20\hat{i} + 30\hat{j})

    Correct
  2. Option B:

    (20i^−30j^)(20\hat{i} - 30\hat{j})

  3. Option C:

    (20i^+45j^)(20\hat{i} + 45\hat{j})

  4. Option D:

    (−4i^+6j^)(-4\hat{i} + 6\hat{j})

Answer: A

Step-by-step solution

E⃗=−∇V=−(10xi^−10yj^)=−10xi^+10yj^\vec{E} = -\nabla V = -(10x\hat{i} -10y\hat{j}) = -10x\hat{i}+10y\hat{j}. At (2,3): −20i^+30j^-20\hat{i}+30\hat{j}.

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
6 April, Evening Shift
Subject
Physics
Chapter
Electrostatics
Topic
Electrostatic Field & motion of charge
The electric potential as a function of x, y is given by V = 5(x 2 … | JEE Main 2026 PYQ with Solution · DhiX AI