Question

Consider an electric field perpendicular to a work bench. When a small charged sphere of mass 3.62 g and charge −19.9 µC is carefully placed in the field, the sphere is in static equilibrium. What are the magnitude and direction of the electric field? (Give the magnitude in N/C.)

Answer 1

Answer:

1782.7 N/C, downward

Explanation:

Since the charged sphere is in static equilibrium, it means that the electric force acting is balanced with the weight of the sphere, so we can write:

$F_G = F_E\\mg = qE$

where

m = 3.62 g = 0.00362 kg is the mass of the sphere

g = 9.8 m/s^2 is the acceleration of gravity

$q=19.9 \mu C= 19.9\cdot 10^{-6}C$ is the magnitude of the charge

E is the magnitude of the electric field

Solving for E,

$E=\frac{mg}{q}=\frac{(0.00362 kg)(9.8 m/s^2)}{19.9\cdot 10^{-6} C}=1782.7 N/C$

In order for the sphere to be in equilibrium, the electric force must be in opposite direction to the weight, so the electric force must point upward. Since the sphere has a negative charge, the electric field has opposite direction to the electric force: so, the electric field direction is downward.