Floating electrons

According to Newton's Second Law, for a body to remain at rest, the net force on it must be 0. There are two forces acting on the electron - (1) Gravitational force = mg (2) Electric force = qE The Gravitational force is acting downward so the electric force must be opposite to it so that the net force is zero. The above two forces must be equal in magnitude but opposite in direction for the electron to remain floating (rest in y direction) Therefore, |mg| = |qE| Where m is the mass of electron, g is acceleration due to gravity, q is charge of electron and E is electric field provided by the infinite sheet of uniform charge density. Using Gauss' Law its possible to prove that $E = \frac {\sigma} {2 \epsilon_0}$ . Solving the equation for $\sigma$ we will find the magnitude of $\sigma$ which comes out to be 1E-21. But note for the force to act upward the charge present on the sheet must repel electron. So, the charge on the sheet must be negative and hence charge density is also negative. That means $\sigma$ = -1E-21

The weight of the electron is its mass multiplied by the acceleration due to gravity

$W = g \times m_e = 8.918 \times 10^{-30} (-a_y) N$

Since the electron floats at rest above the sheet charge, the net force should be zero. The formula for the force on a sheet charge is

$F = q_e\sigma/2\epsilon_0$

$W - F=0$

$\sigma = \frac{2 \times F \times \epsilon_0}{-q_e}$ $= \frac{2 \times 8.918 \times 10^-30 \times \epsilon}{}$ $=-9.8655 \times 10^-22$

The electric field due to a uniformly charged sheet is $\frac{\sigma}{2\epsilon_{o}}$ Since the electron is floating above the sheet and is not moving, we know that $mg =\frac{\sigma}{2\epsilon_{o}}q_{e}$ Solving for $\sigma$ yields $\frac{2mg\epsilon_{o}}{q_{e}} = -9.86\times 10^{-22}$