What must the charge (sign and magnitude) of a particle of mass 1.48 g be for it
ID: 1533603 • Letter: W
Question
What must the charge (sign and magnitude) of a particle of mass 1.48 g be for it to remain stationary when placed in a downward-directed electric field of magnitude 700 N/C ? Use 9.81 m/s2 for the magnitude of the acceleration due to gravity.
What is the magnitude of an electric field in which the electric force on a proton is equal in magnitude to its weight?
Use 1.67×1027 kg for the mass of a proton, 1.60×1019 C for the magnitude of the charge on an electron, and 9.81 m/s2 for the magnitude of the acceleration due to gravity.
Explanation / Answer
1)
Fg = ma
Fg = 0.00148 kg * 9.81 m/s^2
Fg = 0.047 N
The force from the magnetic field needs to be equal to the above, but in the opposite direction (so we need a force
of -0.047 N).
The formula for the magnetic force is:
Fg = qE
-0.047 N = q * 700
q = -6.73 x 10^-5 C
2)
Fg = qE
9.81 * m = q * E
9.81 * (1.67 x 10^27) = (1.60 x 10^19) x E
E = 1.0239 x 10^-7
E = 1.02 x 10^-7
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