A charged conducting sphere with radius 30cm is having a charge of -50nC. A proton is released from rest at R= 10m. Calculate the velocity of the proton when it reaches the surface of the conductor. What happens to the velocity of the proton when the charge on the sphere is changed to +50nC?
V = kq / r
W = qΔv = 1/2mv2
The Attempt at a Solution
First, I connected the potentials for the sphere and the proton:
Vsphere = k(-50nC) / 0.3 = -1500
Vproton = k(+q) / 10 = 1.44e-10
I'm pretty sure this part is where I'm wrong. But I added them up and used -1500 for ΔV.
Then I plugged in +q into W=qΔv equation, solved for v, and got 547722m/s as the answer.
I think I'm close, because it says the correct answer is 52e4 m/s, but I'm not sure where I've messed up.
For the second part where charge on the sphere is +50nC, do I just plug this number into Vsphere and do the same thing to get the velocity?
Any help is appreciated, thanks.