Question

There is a plane of uniform positive charge density \( \sigma \) parallel to the yz-plane and located at \( x = 2d \). A point charge \( q^+ \) is placed at the origin. Solve for the position \( x \) along the x-axis, where a positive test charge will have a net force of zero. 

• A. \( x = \frac{\sqrt{q}}{2 \pi \epsilon_0} \)
• B. No Solution
• C. \( x = \sqrt{\frac{2 \pi \sigma}{d}} \)
• D. \( x = -2d \)

Hint:

When dealing with force balances involving charge densities, use Coulomb’s law and consider the symmetry of the system to find the equilibrium position.

Answer 1

The Correct Option is A

The force on the positive test charge due to a plane of charge is given by the equation: \[ F = \frac{q \sigma}{2 \pi \epsilon_0 x^2} \] For the force to be zero, we need to balance the forces on the test charge. Solving the force equation for \( x \), we find that the position \( x \) at which the net force is zero is: \[ x = \frac{\sqrt{q}}{2 \pi \epsilon_0} \] Hence, the correct answer is (a).