What must the charge (sign and magnitude) of a particle of mass 1.41 g be for it

Question

What must the charge (sign and magnitude) of a particle of mass 1.41 g be for it to remain stationary when placed in a downward-directed electric field of magnitude 660 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 times 10^-27 kg for the mass of a proton, 1.60 times 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

here,

A)

mass , m = 1.41 g = 0.00141 kg

electric feild , E = 660 N/C

let the charge be q

gravitational force = electrostatic force

m * g = q * E

0.00141 * 9.81 = q * 660

q = 2.1 * 10^-5 C

as the electric feild is downward

for upward electric force , the sign of charge must be negative

B)

equaitng the forces

gravitational force = electrostatic force

mp * g= E * e

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

E = 1.02 * 10^-7 N/C

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