MasteringPhysics Chapter 21 part 2-Google Cheome B https//session.masteringphysi

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MasteringPhysics Chapter 21 part 2-Google Cheome B https//session.masteringphysics.com/myct/itemView?assignm 59074710 , PHY 2049C Jerousek Spring 2016 previous 14 of 1 Exercise 21.29 Part A What must the charge (sign and magnitude) of a 1.48-g particle be for il to nemain stationary when placed in a downward-directed electric theld of magnfude 690 N/C? Q- My Answers Gve Part B What is the magnitude of an electric feld in which the electric force on a proton is equal in magnitude to its weight? N/C My Answers gve U Contin . Pine Hils. FLFrom your irter-addess , use precise location Lean more Help Send feedback Pivacy Terms

Explanation / Answer

Part A -
E = 690 N/C
Mass of Particle, m = 1.48 * 10^-3 Kg
Let the charge of the particle = q
Now, For particle to remain stationary in a downward directed electric field.

So,
F = q*E
m*g = q*E
1.48*10^-3 * 9.8 = q* 690
q = 2.1 * 10^-5 C

Charge of the particle, q = - 2.1 * 10^-5 C

Part B
Charge of proton, q = 1.6 * 10^-19 C
Mass of proton, m = 1.67 * 10^-27 kg

F = q*E
m*g = q*E
1.67 * 10^-27 * 9.8 = 1.6 * 10^-19 * E
E = 1.02 * 10^-7 N/C
Magnitude of Electric Field, E = 1.02 * 10^-7 N/C

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