(a) What is the direction and magnitude of an electric field that supports the weight of a free electron near the surface of Earth? (b) Discuss what the small value for this field implies regarding the relative strength of the gravitational and electrostatic forces.

The force exerted on a mass placed in the gravitational field is known as weight. Theexpression for the weight is,

$F_g=mg$

Here, m represents mass of an electron and grepresents acceleration due to earth’s gravity.

When an electron is placed in an electric field, the electrostatic force or Coulomb force acting on the electron is,

F_e = qE

Here, q is the charge on the electron $(q=-1.6\times {10}^{-19}C)$, and E is the strength of the electric field.

To support the weight of the free electron near the surface of Earth, the electrostatic force on the charge due to electric field must be equal to the weight of the electron. Therefore,

F_e= F_g

qE = mg

Here, g is the charge on the electron $\left(q=-1.6\times {10}^{-19}C\right)$ , E is the strength of the electric field, m is the mass of the electron $\left(m=9.1\times {10}^{-31}kg\right)$ , and E is the acceleration due to Earth’s gravitational force near the surface $\left(g=9.81m/s^2\right)$.

The expression for the electric field is,

$E=\frac{mg}{q}$

Substituting all known values,

role="math" localid="1653733373369" $$\begin{gathered} E=\frac{\left(9.1\times {10}^{-31}kg\right)\times \left(9.81m/s^2\right)}{\left(-1.6\times {10}^{-19}C\right)} \\ =-5.58\times {10}^{-11}N/C \end{gathered}$$

Here, negative sign indicates that the electric field is directed downwards.

Hence, the magnitude of an electric field that supports the weight of a free electron near the surface of Earth is $5.58\times {10}^{-11}N/C$and it is directed towards the surface of Earth.

Since a much smaller electric field is needed to create the same magnitude of force as gravity.

Hence, the electrostatic force is much stronger than gravity.