Question

a point charge q=-0.35 nc is fixed at the origin. where must an electron be placed in order for the electric force acting on it to be exactly opposite its weight?

Answer 1

Answer:

238 km above q

Explanation:

Assuming no other external forces present, the electron must be in equilibrium, between gravity (always downward) and the electrostatic force (given by Coulomb's Law) that must be upward.

As the force on the electron due to the negative charge in the origin is repulsive, the electron must be located above the charge q.

In order to find the distance to the origin, we know that the weight and the electrostatic force are equal each other:

$Fe = W\\\frac{k*q*e}{d^{2}} = m*g\\d =\sqrt{\frac{k*q*e}{m*g} } = \sqrt{\frac{9e9 N*m2/C2*0.35e-9C*1.6e-19C}{9.1e-31kg*9.8m/s2} } =237,697 m = 238 km$

⇒ d= 238 Km above q.