(a) Find the charge on each particle. (b) Find the mass of particle 2. kg q 1 =

Question

(a) Find the charge on each particle.

(b) Find the mass of particle 2.
kg

q1 =  C q2 =  C Two particles, with identical positive charges and a separation of 2.65 x 10^-2 m, are released from rest. Immediately after the release, particle 1 has an acceleration vector a1 whose magnitude is 4.20 x 10^3 m/s^2, while particle 2 has an acceleration vector a2 whose magnitude is 8.95 x 10^3 m/s^2. Particle 1 has a mass of6.55 x 10^-6 kg. (a) Find the charge on each particle. (b) Find the mass of particle 2. kg

Explanation / Answer

Note that

m1 = 6.55E-6 kg
a1 = 4.20E3 m/s^2

Thus,

F1 = m1a1 = 0.02751 N

Thus, this is also equal to the Coulomb force,

F = k q1 q2 / r^2

As they are identical charges,

q1 = q2 = Q, and

r = 2.65E-2 m,

k = 8.99E9 N*m^2/kg^2,

Then

0.02751 = (8.99E9)(Q^2)/(2.65E-2)^2

Thus,

Q = 4.64*10^-8 C

for both charges.

q1 = 4.64*10^-8 C   [ANSWER]
q2 = 4.64*10^-8 C   [ANSWER]

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PART B:

By Newton's law of interaction,

F12 = F21

Thus, the force experienced by particle 2 is also 0.02751 N.

As a2 = 8.95E3 m/s^2,

m2 = 0.02751 N/(8.95E3 m/s^2)

m2 = 3.07*10^-6 kg   [ANSWER]

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