Question

Part A
What must the charge (sign and magnitude) of a particle of mass 1.41g be for it to remain stationary when placed in a downward-directed electric field of magnitude 630 N/C?

Use 9.81 m/s^2 for the magnitude of the acceleration due to gravity.

Part B
What is the magnitude of an electric field in which the electric force on a proton is equal in magnitude to its weight?

Use 1.67×10^−27 Kg for the mass of a proton, 1.60×10^−19 C for the magnitude of the charge on an electron, and 9.81 m/s^2 acceleration due to gravity.

Answer 1

A) the charge should be negative so that the force acts upwards

electric force = weight

qE = mg

q*630 = 0.00141*9.8

q = 22*10^-6 C = 22 uC

so the charge is -22 uC

B) qE = mg

1.6*10^-19*E = 1.67*10^-27*9.8

E = 1.02*10^-7 N/C (upwards direction)