Part A:What must the charge (sign and magnitude) of a particle of mass 1.48 g be

Question

Part A: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 650 N/C ?

Use 9.81 m/s2 for the magnitude of the acceleration due to gravity.

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 1.67×1027 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

Part A)

Given that electirc field is downward ,so to remain the particle stationary,the electric Force must be balanced by gravitational force

Electric force = gravitational force

q*E = m*g

q = (m*g)/E = (1.48*10^-3*9.81)/650 = 2.24*10^-5 = 22.4*10^-6 C

sign is negative

so q = -22.4*10^-6 C

Part B) Electric force = weight

q*E = m*g

1.6*10^-19*E = (1.67*10^-27*9.81)

E = 1.02*10^-7 N/C

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