An infinite nonconducting plane has a uniform charge distribution of 4.69 nC/m24.69 nC/m2 and is located in the yz-plane at x = 0. A point charge of 5.01 nC is located at x = 4.33 m. At what position(s) on the x-axis is the electric potential a minimum?  m

The electric potential at a point in space due to a charge distribution is given by the integral of the electric field over the path from a reference point (usually infinity) to the point in question. The electric field due to a uniformly charged infinite plane is constant and points directly away from the plane. The electric field due to a point charge decreases with the square of the distance from the charge and points directly away from the charge.

The electric potential is a minimum where the electric field is zero. This occurs where the electric fields due to the plane and the point charge cancel each other out.

The electric field due to the plane is given by E_plane = σ/2ε0, where σ is the surface charge density and ε0 is the permittivity of free space. The electric field due to the point charge is given by E_point = kQ/r^2, where k is Coulomb's constant, Q is the charge, and r is the distance from the charge.

Setting these two expressions equal to each other gives σ/2ε0 = kQ/r^2. Solving for r gives r = sqrt(kQσ/2ε0).

Substituting the given values gives r = sqrt((8.99 x 10^9 N m^2/C^2)(5.01 x 10^-9 C)(4.69 x 10^-9 C/m^2)/(2 x 8.85 x 10^-12 C^2/N m^2)) = 4.33 m.

Therefore, the electric potential is a minimum at x = 4.33 m.