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.
Show 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.
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).
