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 0.250-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 0.635 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

Using the equation:

d1 = u*t + 0.5*a1*t1^2

u = 0

d1 = 01 m

t1 = 0.400 sec

a1 = 2*d1/t1^2

a1 = 2*1/0.4^2 = 12.5 m/sec^2

F1 = m1*a1 = 0.250*12.5 = 3.125 N

d2 = u*t + 0.5*a2*t2^2

d2 = 1 m

t2 = 0.635 sec

a2 = 2*1/0.635^2 = 4.96 m/sec^2

F2 = 1.24 N

F2 = F1 - qE

qE = F2 - F1

E = (3.125 - 1.24)/(7.65*10^-6)

E = 2.46*10^5 N/C

Now

electric potentia is given by:

V = E*d

d = 1 m

V = 2.46*10^5*1 = 2.46*10^5 V

let me know if you have any doubt.

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