A point charge 4.80 μC is held fixed at the origin. A second point charge 1.40...

A point charge 4.80 μC is held fixed at the origin. A second point charge 1.40 μC with mass of 2.80×10⁻⁴ kg is placed on the x axis, 0.260 m from the origin.
Part A: What is the electric potential energy U of the pair of charges? (Take U to be zero when the charges have infinite separation.) (Answer in Joules)
Part B: The second point charge is released from rest. What is its speed when its distance from the origin is 0.500 m . (Answer in m/s)

Part C: What is its speed when its distance from the origin is 5.00m. (Answer in m/s).

Part D: What is its speed when its distance from the origin is 50.0m. (Answer in m/s).

Solutions

Expert Solution

Part A -

The expression for the electric potential energy = U1 = kQq/r1

So, the electric potential energy=U1 =(9*19^9)[4.80*10^-6][1.40*10^-6] / 0.260 = 0.232 J

Part -B -

The second point charge is released from rest. What is its speed when its distance from the origin is 0.500 m ?

The electric potential energy=U2=kQq/r2 =(9*19^9)[4.80*10^-6][1.40*10^-6] / 0.50 = 0.1210 J

Kinetic energy =U1 - U2= 0.232 - 0.121 = 0.111 J

Now, velocity = v1 = sq rt 2*kinetic energy / mass = sq rt [(2 * 0.111)/ (2.8*10^-4)] = 28.16 m/s

Part C -

The second point charge is released from rest. What is its speed when its distance from the origin is 5.0 m?

The electric potential energy=U3=kQq/r3 =(9*19^9)[4.80*10^-6][1.40*10^-6] / 5.0 = 0.0121 J

Kinetic energy =U1 - U3= 0.232 - 0.0121 = 0.22 J

Therefore, velocity = v2 = sq rt 2*kinetic energy / mass = sq rt (2* 0.22 / 2.8*10^-4) = 39.6 m/s

Part - D

The second point charge is released from rest. What is its speed when its distance from the origin is 50.0 m?

The electric potential energy=U4=kQq/r4 =(9*19^9)[4.80*10^-6][1.40*10^-6]/ 50.0=0.00121 J

Kinetic energy =U1 - U4 = 0.232 - 0.00121 = 0.231 J

Velocity = v3 = sq rt 2*kinetic energy / mass = sq rt (2* 0.231 / 2.8*10^-4) = 40.60 m/s

genius_generous answered 1 year ago

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