Question

What must the charge (sign and magnitude) of a particle of mass 1.45 g be for it to remain stationary when placed in a downward-directed electric field of magnitude 650 N/C Use 9.81 m/s2 for the magnitude of the acceleration due to gravity. View Available Hint(s) Submit 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 1.67x10-27 kg for the mass of a proton, 1.60x101 for the magnitude of the charge on an electron, and 9.81 m/s for the magnitude of the acceleration due to gravity. View Available Hint(s) N/C

Need both parts A and B answered.

Answer 1

We just need to balance the force of gravity with the force from the magnetic field. The gravitational force is equal to

F = m*g
F = 0.00145 kg * 9.81 m/s²
F = 0.0142245 N

The force from the magnetic field needs to be equal to the above, but in the opposite direction

F =q*E

- 0.0142245 N =q*650 N/C

q =-2.18838*10^-5 C

q = -2.188*10^-5 C

B)

m*g =q*E

E = (1.67 * 10^−27 kg)*(9.81 m/s²)/(1.60 * 10^−19 C)

=1.02 * 10^-7 N/C