Mastering Physics Solutions: Suspending Charged Particles Using Electric Fields

Stopping the Proton

Part A = 0.180 m Click to use the calculator/solver for this part of the problem

Solution Below:

Part A

What must the charge (sign and magnitude) of a particle of mass 1.43 g be for it to remain stationary when placed in a downward-directed electric field of magnitude 660 N/C?
Use 9.81 m/s2 for the magnitude of the acceleration due to gravity.

For this problem, you just need to balance the force of gravity with the force from the magnetic field. The gravitational force is equal to (make sure to convert the mass of the particle to kilograms):

Fg = ma
Fg = 0.00143 kg * 9.81 m/s2
Fg = 0.01403 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 -0.0140 N). The formula for the magnetic force is:

Fg = qE

We know the required force, and the magnitude of the field was already given:

-0.01403 N = q * 660
q = -2.125 * 10^-5 C
q = -2.13 * 10^-5 C

-2.13 * 10^-5 C

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 for the magnitude of the acceleration due to gravity.

To solve this part, just use the same concept as in Part A:

Fg = qE
9.81 * m = q * E
9.81 * (1.67 * 10^−27) = (1.60 * 10^−19) * E
E = 1.0239 * 10^-7
E = 1.02 * 10^-7

E = 1.02 * 10^-7

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