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 0.220-kg ball with zero net charge was dropped from rest at a height of 1.00 m. The ball landed 0.400 s later. Next, the ball was given a net charge of 7.65 C and dropped in the same way from the same height. This time the ball fell for 0.630 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.)

Explanation / Answer

Let us first find the acceleration without the charge,

h= vi*t -1/2gt1^2

-1.0 = 0* -1/2*awithout*0.4^2   => awithout = 12.5 m/s^2

Now find the acceleration with the charge,

h= vi*t -1/2gt2^2

-1.0 = 0* -1/2*awith*0.63^2   => awith = 5.04 m/s^2

acharge = awith - awittout = 5.04 - 12.5 = -7.46 m/s^2

Now velocity at the ground with

vf^2 = vi^2 +2ah

vf^2 = 0^2 +2*-7.46*-1.0   => vf= 3.9 m/s

KE = W = -qV

1/2mv^2 = - qV

1/2*(0.220)*(3.9)^2 = -(7.65*10^-6)*V   => V = -2.19*10^5 V

V = Vf - Vi

2.19*10^5 = -2.19*10^5 – 0    => Vf = -2.19*10^5 V

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