Question

When released from rest in a uniform electric field of magnitude 1.00 x 10^4 N/C, a charged particle travels 2.00 cm in 2.00 x 10A-7s in the direction of the field. You can ignore any nonelectric forces acting on the particle. Set up your solution to answer the following questions: (i) Is the particle positively or negatively charged? (ii) What is the charge-to-mass ratio of this particle, i.e. the ratio of the charge to the mass? (ii) Other experiments show that the mass of this particle 6.64 x 10-27 kg. What is the charge of the particle? Apply the following steps in your solution: a) Describe the approach you will take b) Visualize the problem c) Set up the equations you will need to solve the problem d) Apply algebraic techniques to solve for the unknown of interest e) Calculate the answers to parts (i)- (ii)

Answer 1

i)

a positive charge particle in an electric field experience force in same direction as the electric field.

a negative charge particle in an electric field experience force in opposite direction as the electric field.

Since the charged particle experience force in the direction of electric field, hence charged particle is positive in nature.

Positively charged.

ii)

q = magnitude of charge on charged particle

m = mass of the charged particle

a = acceleration

x = distance traveled in the direction of electric field = 2 cm = 0.02 m

t = time of travel = 2 x 10^-7 s

v_{o =} initial velocity = 0 m/s (Since it starts from rest)

using the kinematics equation

x = v_o t + (0.5) a t²

0.02 = 0 t + (0.5) a (2 x 10^-7)²

a = 1 x 10¹² m/s²

E = magnitude of electric field = 1 x 10⁴ N/C

Net force on the particle is due to the electric field which is given as

F_e = q E

also , from newton;'s second law

F_e = ma

hence

ma = q E

q/m = a/E

q/m = (1 x 10¹²)/(1 x 10⁴)

q/m = 1 x 10⁸

iii)

m = mass of the particle = 6.64 x 10^-27 kg

q/m = 1 x 10⁸

q/(6.64 x 10^-27) = 1 x 10⁸

q = 6.64 x 10^-19 C