# A point charge 4.80 μC is held

A point charge 4.80 μC is held fixed at the origin. Asecond point charge 1.40 μC with mass of2.80×10−4 kg is placed on the x axis, 0.260 mfrom the origin.Part A: What is the electric potential energy U of thepair of charges? (Take U to be zero when the charges haveinfinite separation.) (Answer in Joules)Part B: The second point charge is released from rest. What is itsspeed when its distance from the origin is 0.500 m . (Answer inm/s)

Part C: What is its speed when its distance from the origin is5.00m. (Answer in m/s).

Part D: What is its speed when its distance from the origin is50.0m. (Answer in m/s).

Answer:

Part A –

The expression for the electric potential energy = U1 =kQq/r1So, 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 speedwhen 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 JKinetic energy =U1 – U2= 0.232 – 0.121 = 0.111 JNow, 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 speedwhen 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 JKinetic energy =U1 – U3= 0.232 – 0.0121 = 0.22 JTherefore, 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 speedwhen 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 JKinetic energy =U1 – U4 = 0.232 – 0.00121 = 0.231 JVelocity = v3 = sq rt 2*kinetic energy / mass = sq rt (2* 0.231 /2.8*10^-4) = 40.60 m/s

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