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 0.350 s later. Next, the ball was given a net charge of 7.70 μC and dropped in the same way from the same height. This time the ball fell for 0.650 s before landing. A.What is the electric potential at a height of 1.00m above the ground on this planet, given that the electric potential at ground level is zero? (Air resistance can be ignored.)

Answer 1

Answer:

$\Delta V=316167V$

Explanation:

The difference of electric potential between two points is given by the formula $\Delta V=Ed$, where d is the distance between them and E the electric field in that region, assuming it's constant.

The electric field formula is $E=\frac{F}{q}$, where F is the force experimented by a charge q placed in it.

Putting this together we have $\Delta V=\frac{Fd}{q}$, so we need to obtain the electric force the charged ball is experimenting.

On the second drop, the ball takes more time to reach the ground, this means that the electric force is opposite to its weight W, giving a net force $N=W-F$. On the first drop only W acts, while on the second drop is N that acts.

Using the equation for accelerated motion (departing from rest) $d=\frac{at^2}{2}$, so we can get the accelerations for each drop (1 and 2) and relate them to the forces by writting:

$a_1=\frac{2d}{t_1^2}$

$a_2=\frac{2d}{t_2^2}$

These relate with the forces by Newton's 2nd Law:

$W=ma_1$

$N=ma_2$

Putting all together:

$N=W-F=ma_1-F=ma_2$

Which means:

$F=ma_1-ma_2=m(a_1-a_2)=m(\frac{2d}{t_1^2}-\frac{2d}{t_2^2})=2md(\frac{1}{t_1^2}-\frac{1}{t_2^2})$

And finally we substitute:

$\Delta V=\frac{Fd}{q}=\frac{2md^2}{q}(\frac{1}{t_1^2}-\frac{1}{t_2^2})$

Which for our values means:

$\Delta V=\frac{2(0.21Kg)(1m)^2}{7.7\times10^{-6}C}(\frac{1}{(0.35s)^2}-\frac{1}{(0.65s)^2})=316167V$