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Electric Potential

 
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keocat



Joined: 13 Mar 2011
Posts: 1
Location: canada

PostPosted: Sun Mar 13, 2011 11:49 pm    Post subject: Electric Potential Reply with quote

Alright, so I don't really know where to start this problem. I know that there must be some comparison between the two situations, but I don't kno what or where to start to figure it out!

Question:

Long, long ago, on a planet far, far away, a physics experiment was carried out. First, a 0.240 kg ball with zero net charge was dropped from rest at a height of 1.00 m. The ball landed 0.350 s later. Next, the ball was given a net charge of 7.80 \mu C
and dropped in the same way from the same height. This time the ball fell for 0.650 s before landing.

What is the electric potential at a height of 1.00 m above the ground on this planet, given that the electric potential at ground level is zero? (Air resistance can be ignored.)

I believe that V = kq/r or V = qU

Again, I'm not sure how I should be looking at this problem, since I seem to be missing a lot of information.
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hepcj
Site Admin


Joined: 23 Jun 2007
Posts: 125

PostPosted: Tue Mar 22, 2011 11:40 pm    Post subject: Reply with quote

First we need to calculate the force due to gravity when there is no charge on the object. We can use the equations of linear motion.

s = u t + 0.5 g t^2

1 = 0.5 g 0.35^2

which gives g = 16.3265 m/s^2

calculate the acceleration on the object when the charge is added. This will the sum of the force due to gravity and the force in the opposite direction due to the repulsive charge. (it has to act in the opposite direction because the time taken to fall is longer.)

So we have

F = m g h - q E
or

a = m g - q E/m

substitute this expression for acceleration into the equation of motion where the charged object falls:

1 == 0.5 a 0.65^2

and solve for E and it should be 3.57 x 10^5 N/C

from V = E d
3.57 x 10^5 V relative to the zero of potential.
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