JEE Main 2025 Apr 2 Shift 1, Physics Q7: Coulomb's Law And Electric Field

JEE Main2025Apr 2, Shift 1Physics

A point charge $+ q$ is placed at the origin. A second point charge $+ 9 q$ is placed at $(d, 0, 0)$ in Cartesian coordinate system. The point in between them where the electric field vanishes is :

A
$(4 d /3, 0, 0)$
B
$(d /4, 0, 0)$
C
$(3 d /4, 0, 0)$
D
$(d /3, 0, 0)$

Solution

Let $P = (x, 0, 0)$ be a point between the two charges with $0 < x < d$ . The fields from $+ q$ and $+ 9 q$ point in opposite directions along the x-axis at $P$ , so for the net field to vanish their magnitudes must be equal:

$\frac{k q}{x ^{2}} = \frac{k ( 9 q )}{( d - x ) ^{2}}$

Take square roots (both sides positive):

$\frac{1}{x} = \frac{3}{d - x} \Rightarrow d - x = 3 x \Rightarrow x = \frac{d}{4}$

$∴$ The null point is at $(\frac{d}{4}, 0, 0)$ , which is closer to the smaller charge — as expected. Option (2) is correct.

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JEE Main 2025 Apr 2 Shift 1, Physics Q7: Coulomb's Law And Electric Field

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