Physics Q39 – JEE-MAIN 2025

Home Physics Electric Charges and FieldsQuestion #112

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 :

Clue 1 of 3: Principle of Superposition

The net electric field at any point is the vector sum of the electric fields due to individual charges.

🥷

Ninja Strategy•Proximity to Smaller Charge

For two positive charges, the null point (where the electric field is zero) is always closer to the smaller magnitude charge. Since $+ q$ is smaller than $+ 9 q$ , the null point must be closer to the origin than to $d$ . Option $(d / 4, 0, 0)$ is the only one that is closer to the origin than to $d / 2$ (midpoint).

🧠 Full Solution Path

● Video Walkthrough

Step 1: Identify Charges and Positions✦ Active

A charge $Q_{1} = + q$ is placed at the origin $(0, 0, 0)$ . A second charge $Q_{2} = + 9 q$ is placed at $(d, 0, 0)$ . We are looking for a point $(x, 0, 0)$ between them where the net electric field is zero. This implies $0 < x < d$ .

Step 2: Apply Condition for Zero Electric Field○ Expand

The electric field due to $Q_{1}$ at point $x$ is $E_{1} = k \frac{q}{x^{2}}$ (directed along the positive x-axis). The electric field due to $Q_{2}$ at point $x$ is $E_{2} = k \frac{9 q}{(d - x)^{2}}$ (directed along the negative x-axis). For the net electric field to be zero, their magnitudes must be equal:

E_{1} = E_{2} ⟹ k \frac{q}{x^{2}} = k \frac{9 q}{(d - x)^{2}}

💡 Teacher's Secret Hint

Remember that electric fields due to positive charges point away from the charge.

Step 3: Solve for the Position x○ Expand

Cancel $k$ and $q$ from both sides and simplify the equation:

\frac{1}{x^{2}} = \frac{9}{(d - x)^{2}}

Taking the square root of both sides (since $x$ and $d - x$ are positive in the region between the charges):

\frac{1}{x} = \frac{3}{d - x}

Cross-multiply to solve for $x$ :

d - x = 3 x ⟹ d = 4 x ⟹ x = \frac{d}{4}

Thus, the point where the electric field vanishes is $(d / 4, 0, 0)$ .

💡 Teacher's Secret Hint

Ensure you consider only the positive square root, as the point must be physically located between the charges.

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Physics Question 39 – JEE-MAIN 2025

Clue 1 of 3: Principle of Superposition

🧠 Full Solution Path