Question

Two equally charged particles are held 4.1 ✕ 10⁻³ m apart and then released from rest. The initial acceleration of the first particle is observed to be 7.9 m/s² and that of the second to be 9.2 m/s². The mass of the first particle is 6.3 ✕ 10⁻⁷ kg.

(a) What is the mass of the second particle?
____________kg

(b) What is the magnitude of the charge of each particle?
_____________ C

Answer 1

Using Newton's 2nd law on 1st particle:

Fnet = m1*a1

Fnet = 6.3*10^-7 kg*7.9 m/sec^2 = 4.977*10^-6 N

Now from newton's 3rd law the same force would be applied on 2nd particle.

As every action has equal and opposite reaction So Net force on particle 2:

Fnet = m2*a2

m2 = Fnet/a2

m2 = 4.977*10^-6 N/(9.2 m/sec^2)

m2 = 5.4*10^-7 kg = mass of 2nd particle

Part B.

Now Net electrostatic force between them would also be equal to total force on each particle

Fnet = Fe

Fnet = k*q1*q2/r^2

q1 = q2 = Q

r = distance between both particles = 4.1*10^-3 m

k = electrostatic constant = 9*10^9

So,

Fnet = k*Q^2/r^2

Q = sqrt (Fnet*r^2/k)

Using known values:

Q = sqrt (4.977*10^-6*(4.1*10^-3)^2/(9*10^9))

Q = 9.6*10^-11 C = Charge on each particle

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