What must the charge (sign and magnitude) of a particle of mass 1.47 g be for it

Question

What must the charge (sign and magnitude) of a particle of mass 1.47 g be for it to remain stationary when placed in a downward-directed electric field of magnitude 700N/C?

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

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

Use 1.67x10^-27 kg for the mass of a proton, 1.60x10^-19 C for the magnitude of the charge on an electron, and 9.81m/s^2 for the magnitude of the acceleration due to gravity.

Explanation / Answer

a) m = 1.47 * 10^-3 kg
Take downward direction as positive.
Fg = m g
E = 700 N/C
Fe = q E
Fe + Fg = 0
q E + m g = 0
q = -m g/E = -1.47 * 10^-3 * 9.81/700 = -2.06 * 10^-5 C
Ans: -2.06 * 10^-5 C


b) q E= m g
E = m g/q = 1.67 * 10^-27 * 9.81/(1.60 * 10^-19) = 1.03 * 10^-7 N/C
Ans: 1.03 * 10^-7 N/C

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