00:01
So in this question we have a conducting sphere of radius r equals two centimeters, and it's charged with a charge q equals 10 nanoculoms.
00:15
So in part a, we want to calculate the electric potential at the surface of the sphere.
00:20
And this is going to behave as if we had a point charge at the center of the sphere.
00:25
So the potential at the surface, so two centimeters, is going to be q.
00:32
Over 4 pi epsilon nought r, which is 10 nanocolon, so 10 times 10 to the minus 9, divided by 4 pi times 8 .85 times 10 to the minus 12, times the radius, which is 0 .02 meters.
00:53
And we get a potential of 4 ,500 volts.
01:01
More precisely it's 4496 volts to four significant figures.
01:09
So now we want to calculate the electric potential at a distance of v at 6 centimetres.
01:18
Well, this is q over 4 pi epsilon nought times 3r, which is 1 third of v at 2 centimeters.
01:28
So all we need to do is divide what we got before by 3.
01:31
And we get 1 ,500 volts, all to three significant figures, or four significant figures.
01:38
This is 4 -4 -4 -4 -4 -4 -4 -4 -9 volts.
01:47
So now we release a proton at the surface of the sphere, and it accelerates because it's being repulsed.
01:55
So we want to use conservation of energy to get the speed of the proton when it reaches six centimeters.
02:05
So delta k is going to be minus delta u for potential energy...