A tiny ball (mass = 0.015 kg) carries a charge of -19C. What is the electric field magnitude that would cause the ball to float above the ground?
E*Q = M*g
Solve for E, in units of Newtons/Coulomb (same as Volts/meter)
To find the electric field magnitude that would cause the ball to float above the ground, we need to consider the gravitational force acting on the ball and the electric force.
The electrical force experienced by the ball can be calculated using the formula:
F_electric = q * E
where F_electric is the electrical force, q is the charge of the ball, and E is the electric field.
The gravitational force acting on the ball can be calculated using the formula:
F_gravity = m * g
where F_gravity is the gravitational force, m is the mass of the ball, and g is the acceleration due to gravity.
For the ball to float, the net force acting on the ball should be zero. Therefore, the electrical force (F_electric) should be equal in magnitude and opposite in direction to the gravitational force (F_gravity). Mathematically, we can express this as:
|F_electric| = |F_gravity|
Substituting the formulas for the electrical and gravitational forces, we have:
|q * E| = |m * g|
Since the ball has a negative charge, the electric field (E) should be in the opposite direction to the gravitational force (g). So, the magnitude of the electric field required to counteract the gravitational force is given by:
E = (m * g) / |q|
Plugging in the values given in the problem:
m = 0.015 kg
g = 9.8 m/s^2
q = -19C
E = (0.015 kg * 9.8 m/s^2) / |-19C|
E = 0.147 m/s^2/ C
Therefore, the electric field magnitude required for the ball to float above the ground is 0.147 m/s^2/ C.