What must the charge (sign and magnitude) of a particle of mass 1.46 g be for it

Question

What must the charge (sign and magnitude) of a particle of mass 1.46 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.

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

F = qE q = F/E = (1.46*10^-3) *9.8 / 600 =2.384*10^-5 F=qE E = F/q given F = m = 1.67×10-27 *9.8kg so E =1.67×10^-27 *9.8kg / 1.60×10^-19 C = 10.22*10^-8 N/C ANSWER

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