JEE Main 2026 — Jan 22, Shift 1 — Physics solution

JEE Main2026Jan 22, Shift 1Physics

Electric field in a region is given by $E = A x \hat{i} + B y \hat{j}$ , where $A = 10$ V/m $^{2}$ and $B = 5$ V/m $^{2}$ . If the electric potential at a point (10, 20) is 500 V, then the electric potential at origin is ______ V.

A
1000
B
500
C
2000
D
0

Solution

Use $V_{f} - V_{i} = - \int_{i}^{f} E \cdot d r$ . Take the straight path from origin (i) to $(10, 20)$ (f):

$V (10, 20) - V (0, 0) = - \int_{(0, 0)}^{(10, 20)} (10 x \hat{i} + 5 y \hat{j}) \cdot (d x \hat{i} + d y \hat{j}) = - [5 x^{2} + \frac{5 y ^{2}}{2}]_{(0, 0)}^{(10, 20)}$ .

$= - [5 (100) + \frac{5 ( 400 )}{2}] = - [500 + 1000] = - 1500$ .

So $500 - V_{0} = - 1500$ , giving $V_{0} = 2000$ V.

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