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 620 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

Here ,

part a) let the charge is q

as the electric field is downwards for electric force to be upwards

the charge must be negative

m * g = q * E

-1.48 *10^-3 * 9.8 = 620 * q

q = -2.34 *10^-5 C

the charge on the on the object must be -2.34 *10^-5 C

b)

let the electric field is E

as m * g = e * E

E = 1.67 *10^-27 * 9.8/(1.602 * 10^-19)

E = 1.02 *10^-7 N/C

the electric field is 1.02 *10^-7 N/C

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects