Two point charges are fixed 4.0cm apart from each other. Their charges are Q1=Q2

Question

Two point charges are fixed 4.0cm apart from each other. Their charges are Q1=Q2 = 6.5 C , and their masses are m1 = 1.5 mg and m2 = 3.5 mg.

Part A

If Q1 is released from rest, what will be its speed after a very long time?

Part B

If both charges are released from rest at the same time, what will be the speed of Q1 after a very long time?

A +20 C point charge is placed 48 cm from an identical +20 C charge.

Part A

How much work would be required for an external force to move a +0.40 C test charge from a point midway between them to a point 13 cm closer to either of the charges?

Explanation / Answer

Part A, from the conservation of the momentum, since the initial momentum is zero the final momentum must be zero,

suppose the velocity of mass m1 is v1 and mass m2 is v2.

then m1v1 = m2v2 = p( velocities are in opposite direction)

So the final kinetic energy = 0.5(m1*v1^2+m2*v2^2) = 0.5*(m1 v^2 + m1^2 v^2/m2)

The initial potential energy will convert into the final kinetic energy.

0.5*m1*v^2(1 + m1/m2) = kq1q2/d

0.5m1v^2 = kq1q2/d(1 + m1/m2) = 9*10^9*6.5^2*10^-12/(0.04*(1+1.5/3.5)) = 6.65 J

Part B:

Speed of Q1, v = sqrt(6.65*2/(1.5*10^-3)) = 94.16 m/s

2)

A +20 C point charge is placed 48 cm from an identical +20 C charge.

The work done required will be equal to the change in potential energy of the system.

work done = 0.4*10^-6(9*10^9*20*10^-6(1/0.24-1/0.35)+9*10^9*20*10^-6*(1/0.24-1/0.13)) = -0.159 J

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