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(1) What must the charge (sign and magnitude) of a particle of mass 1.50 g be fo

ID: 2005878 • Letter: #

Question

(1) What must the charge (sign and magnitude) of a particle of mass 1.50 g be for it to remain stationary when placed in a downward-directed electric field of magnitude 600 N/C?
Use 9.81 m/s^2 for the magnitude of the acceleration due to gravity.

(2) 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.

Explanation / Answer

(1).mass m = 1.5 g = 1.5 * 10 ^-3 kg electric field E = 600 N / C Let the charge of a particle be q and it is negative In equilibrium position , weight = force due to electric field mg = Eq from this charge q = mg / E = -2.45 * 10 ^-5 C Since it is negative (2). the electric force on a proton = magnitude of weight of proton Eq = mg where mass of proton m = 1.67×10-27 kg charge of proton q = 1.60×10-19 C the magnitude of the acceleration due to gravity g = 9.81 m / s^ 2 from above equation E = mg / q = 1.022 *10^-7 N / C