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 0.350 s later. Next, the ball was given a net charge of 7.60 muC and dropped in the same way from the same height. This time the ball fell for 0.645 s before landing.

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

No one has answered this question yet.

To answer this question, you need to understand the relationship between electric potential energy and gravitational potential energy.

The electric potential energy, denoted as PE_electric, is given by the equation:

PE_electric = q * V,

where q is the charge and V is the electric potential.

Now, let's start by finding the electric potential at the height of 1.00 m above the ground. Given that the electric potential at ground level is zero, we can assume that the potential energy difference between the ground and 1.00 m above the ground is equal to the electric potential at that height.

First, let's calculate the initial kinetic energy (K1) of the ball when it was dropped from rest at a height of 1.00 m. We can use the equation:

K1 = (1/2) * m * v1^2,

where m is the mass of the ball and v1 is the velocity at the time it reaches the ground.

Using the equation, v1 = g * t1 (where g is the acceleration due to gravity and t1 is the time taken to fall), we can rewrite the equation as:

K1 = (1/2) * m * (g * t1)^2.

Next, let's calculate the final kinetic energy (K2) of the ball when it fell for 0.350 s. Now there is an additional force acting on the ball due to the electric potential energy:

K2 = (1/2) * m * v2^2 + PE_electric.

Using the equation, v2 = g * t2, we can rewrite the equation as:

K2 = (1/2) * m * (g * t2)^2 + PE_electric.

Since the ball is given a net charge of 7.60 muC, we can convert it to Coulombs: 7.60 muC = 7.60e-6 C.

Now, the change in electric potential energy is given by the equation:

∆PE_electric = q * ∆V,

where ∆V is the change in electric potential.

As we know that the ball is dropped from the same height as before, its change in potential energy will remain the same. Therefore, we have:

∆PE_electric = PE_electric.

Now, let's put all the equations together:

K2 = (1/2) * m * (g * t2)^2 + PE_electric,

K1 = (1/2) * m * (g * t1)^2.

Since both equations involve the sum of gravitational potential energy and electric potential energy, we can subtract K1 from K2 to eliminate the kinetic energy terms:

K2 - K1 = (1/2) * m * (g * t2)^2 - (1/2) * m * (g * t1)^2,

(1/2) * m * (g * t2)^2 - (1/2) * m * (g * t1)^2 = PE_electric,

(g * t2)^2 - (g * t1)^2 = PE_electric.

Now, we can substitute the given values into the equation:

(9.8 m/s^2 * 0.645 s)^2 - (9.8 m/s^2 * 0.350 s)^2 = PE_electric,

(9.8 * 0.645)^2 - (9.8 * 0.350)^2 = PE_electric.

Finally, we can solve the equation to calculate the electric potential at a height of 1.00 m above the ground on this planet.