Question

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

• A. \( \left(\frac{4d}{3}, 0, 0\right) \)
• B. \( \left(\frac{d}{4}, 0, 0\right) \)
• C. \( \left(\frac{3d}{4}, 0, 0\right) \)
• D. \( \left(\frac{d}{3}, 0, 0\right) \)

Hint:

In problems involving multiple charges, use the principle of superposition for electric fields and set the total field equal to zero to find the point of cancellation.

Answer 1

The Correct Option is B

Let the electric field at point \( P \) in between the charges be zero. Let the position of \( P \) be at a distance \( x \) from the origin, where the electric field due to both charges cancels each other.  
The electric field due to a point charge is given by: \[ E = \frac{kq}{r^2} \] For the electric field to be zero at point \( P \), the fields due to both charges must be equal and opposite. So: \[ \frac{kq}{x^2} = \frac{k(9q)}{(d - x)^2} \] Simplifying: \[ \frac{1}{x^2} = \frac{9}{(d - x)^2} \] Solving for \( x \): \[ d - x = 3x \quad \Rightarrow \quad d = 4x \quad \Rightarrow \quad x = \frac{d}{4} \] Thus, the coordinate of point \( P \) is \( \left(\frac{d}{4}, 0, 0\right) \).