Q67 / 75 · JEE Main 2026 (22 January Shift 1) · Physics
Electric field in a region is given by $\vec{E}=A x \hat{i}+B y \hat{j}$, where $A=10 \mathrm{~V} / \mathrm{m}^{2}$ and $B=5 \mathrm{~V} / \mathrm{m}^{2}$. If the electric potential at a point $(10,20)$ is 500 V, then the electric potential at origin is $\_\_\_\_$ V.
- A0
- Correct answerB2000
- C500
- D1000
Solution
The electric field is related to potential by $\vec{E} = -\nabla V$.
Given: $\vec{E} = Ax\hat{i} + By\hat{j}$ with $A = 10$ V/m² and $B = 5$ V/m²
This means:
$\frac{\partial V}{\partial x} = -Ax$ and $\frac{\partial V}{\partial y} = -By$
Integrating:
$V(x,y) = -\frac{Ax^2}{2} - \frac{By^2}{2} + C = -5x^2 - \frac{5y^2}{2} + C$
Using the boundary condition at $(10, 20)$ where $V = 500$ V:
$500 = -5(100) - \frac{5(400)}{2} + C$
$500 = -500 - 1000 + C$
$C = 2000$ V
At the origin $(0, 0)$:
$V(0,0) = -5(0) - \frac{5(0)}{2} + 2000 = 2000$ V
Given: $\vec{E} = Ax\hat{i} + By\hat{j}$ with $A = 10$ V/m² and $B = 5$ V/m²
This means:
$\frac{\partial V}{\partial x} = -Ax$ and $\frac{\partial V}{\partial y} = -By$
Integrating:
$V(x,y) = -\frac{Ax^2}{2} - \frac{By^2}{2} + C = -5x^2 - \frac{5y^2}{2} + C$
Using the boundary condition at $(10, 20)$ where $V = 500$ V:
$500 = -5(100) - \frac{5(400)}{2} + C$
$500 = -500 - 1000 + C$
$C = 2000$ V
At the origin $(0, 0)$:
$V(0,0) = -5(0) - \frac{5(0)}{2} + 2000 = 2000$ V
Hence, the correct answer is option: 'B'.