To find the charge needed for the baseball to float, we can use the equation F = qE, where F is the force of the electric field, q is the charge, and E is the electric field strength.
Given:
Electric field strength (E) = 150 N/C
Mass of the baseball (m) = 145 g = 0.145 kg
Acceleration due to gravity (g) = 9.8 m/s^2
Since the baseball is weightless when floating, the electrical force acting on it will balance the force of gravity:
qE = mg
q(150) = (0.145)(9.8)
q = 0.145(9.8) / 150
q = 0.00949 C
Therefore, the required charge for the baseball to float as though it were weightless is 0.00949 C.
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 145 g baseball need to float as though it were weightless?
Give your answer for the magnitude of the required charge in C:
1 answer