One particle has a mass of 4.84 x 10-3 kg and a charge of +6.47 C. A second part

Question

One particle has a mass of 4.84 x 10-3 kg and a charge of +6.47 C. A second particle has a mass of 7.70 x 10-3 kg and the same charge. The two particles are initially held in place and then released. The particles fly apart, and when the separation between them is 0.140 m, the speed of the 4.84 x 10-3 kg-particle is 141 m/s. Find the initial separation between the particles.

Two equipotential surfaces surround a +3.70 x 10-8-C point charge. How far is the 200-V surface from the 44.0-V surface?

Explanation / Answer

1. particle gain momentum due to electrostatic force.

that is internal force for two charge system.

Applying momentum conservation,

initial = final momentum

0 = (4.84 x 10^-3)(141) - (7.70 x 10^-3)(v)

v = 88.63 m/s

Now applying energy conservation,

PEi + KEi = PEf + KEf

k q^2 / r + 0 = k q^2 / 0.140 + [ (4.84 x 10^-3)(141^2)/2 + (7.70 x 10^-3)(88.63^2) / 2]

{ k q^2 = (9 x 10^9)(6.47 x 10^-6)^2 = 0.377}

0.377 / r = 0.377 / 0.140 + 78.35

r = 4.65 x 10^-3 m Or 4.65 mm .....Ans

-------------------

2. V = k Q / R

200 = (9 x 10^9)(3.70 x 10^-8) / R1

R1 = 1.665 m

44 = (9 x 10^9)(3.70 x 10^-8) / R2

R2 = 7.57 m

d = R2 - R1 = 5.90 m

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