Due to electric effects on the earth and in the atmosphere, there is an electric field near the earth's surface that points down. Assuming the electric field is uniform with a magnitude of 150 N/C, what charge would a 145g baseball need to float as though it were weightless?
Give your answer for the magnitude of the required charge in C:
To find the magnitude of the required charge in coulombs (C), we can use the equation:
F = qE
Where:
F is the force experienced by the charged object (baseball)
q is the charge of the object
E is the electric field intensity
In this case, we want the baseball to float as though it were weightless, meaning the gravitational force on the baseball is balanced by the electrical force:
F_gravity = F_electric
The gravitational force acting on the baseball is given by:
F_gravity = m * g
Where:
m is the mass of the baseball
g is the acceleration due to gravity (approximately 9.8 m/s^2)
The electrical force experienced by the baseball is given by:
F_electric = q * E
Setting the two forces equal to each other:
m * g = q * E
Now we can solve for q:
q = (m * g) / E
Substituting the given values:
m = 145g = 0.145 kg
g = 9.8 m/s^2
E = 150 N/C
q = (0.145 kg * 9.8 m/s^2) / 150 N/C
q ≈ 0.00941 C
Therefore, the magnitude of the required charge for the baseball to float as though it were weightless is approximately 0.00941 coulombs.