Long, long ago, on a planet far, far away, a physics experiment was carried out.

Question

Long, long ago, on a planet far, far away, a physics experiment was carried out. First, a 0.210-kg ball with zero net charge was dropped from rest at a height of 1.00 m. The ball landed 0.450s later. Next, the ball was given a net charge of 7.80?C
and dropped in the same way from the same height. This time the ball fell for 0.635s 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.)

Actual answer please

Explanation / Answer

Use d = 1/2 a*t^2
We have d =1 and we have t = 0.45
a = 2/0.45^2 = 9.8765m/s^2
So the gravitational force on the ball is ma = 0.21*9.8765 = 2.07N
When the ball is charged, the electric force = qE apparently opposes the gravitational force because it takes the ball longer to descend. Fnet = 2.07 - qE = 0.21*a1
1 = 1/2*a1*0.635^2 => a1 = 4.96 => Fnet = m*a1 = 0.21*a1 = 1.04N
That means the opposing force due to the electric field is 2.07-1.04 = 1.02N = qE
E = 130769 V/m
At 1m V=130769 Volts
This is a very strong field which is expected since it actually effects the rate of fall.

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