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.)
To find the electric potential at a height of 1.00 m above the ground, we need to consider the change in potential energy of the charged ball as it falls.
First, let's calculate the gravitational potential energy of the uncharged ball when it falls from a height of 1.00 m. The formula for gravitational potential energy is:
PE_gravity = m * g * h
Where,
PE_gravity is the gravitational potential energy,
m is the mass of the ball,
g is the acceleration due to gravity, and
h is the height.
Given:
m = 0.250 kg (mass of the ball)
h = 1.00 m (height)
We need to determine the acceleration due to gravity. The time the ball takes to fall can be used to calculate this.
The equation for the time it takes for an object to fall freely from height h is:
h = (1/2) * g * t^2
Rearranging the equation to solve for g:
g = (2h) / t^2
Given:
h = 1.00 m
t = 0.350 s (time taken by the uncharged ball)
Now we can calculate the acceleration due to gravity:
g = (2 * 1.00 m) / (0.350 s)^2
After calculating, we get:
g = 8.979 m/s^2
Substitute this back into the formula for gravitational potential energy:
PE_gravity = 0.250 kg * 8.979 m/s^2 * 1.00 m
PE_gravity ≈ 2.245 J
Next, let's calculate the gravitational potential energy of the charged ball using the same approach.
We know that the charge of the ball is 7.60 muC (microCoulombs), which can be converted to Coulombs:
Charge = 7.60 muC = 7.60 x 10^-6 C
The formula for electric potential energy is:
PE_electric = charge * electric potential
We need to find the electric potential. Given that the electric potential at ground level is zero, we can find the electric potential at a height of 1.00 m by calculating the difference in electric potential energy between the two heights.
PE_electric_at_1m - PE_electric_at_ground = ΔPE_electric
Given that PE_electric_at_ground = 0, we have:
ΔPE_electric = PE_electric_at_1m - 0
Now, let's calculate ΔPE_electric.
Given:
charge = 7.60 x 10^-6 C
t = 0.645 s (time taken by the charged ball)
We can find the electric potential energy of the charged ball at a height of 1.00 m using the formula:
PE_electric = charge * electric potential
electric potential = PE_electric / charge
electric potential = (ΔPE_electric + 0) / charge
electric potential = ΔPE_electric / charge
Substitute the values:
electric potential = ((0.250 kg * 8.979 m/s^2 * 1.00 m) - 0) / (7.60 x 10^-6 C)
After calculation, we get:
electric potential ≈ 74,331 V
So, the electric potential at a height of 1.00 m above the ground on this planet is approximately 74,331 V.