two identical 200 gram balls of diameter 3.0 cm are 5.0 cm apart. each carries a charge of 6.5 micro coulombs. one of the balls is realesed. find its acceleration if gravitational force is negligible small?
F = k q1 q2 /r^2
a = F/m = k q1 q2/(m r^2)
To find the acceleration of the released ball, we need to consider the electrostatic force between the two charged balls. The formula to calculate the electrostatic force is given by Coulomb's law:
F = k * (q1 * q2) / r²
Where:
F is the electrostatic force between the two balls
k is the electrostatic constant (8.99 x 10^9 N m²/C²)
q1 and q2 are the charges on the two balls (6.5 x 10^-6 C, since they are identical)
r is the distance between the centers of the two balls (5.0 cm + 3.0 cm/2 + 3.0 cm/2 = 6.5 cm = 0.065 m)
Now we can calculate the electrostatic force between the balls:
F = (8.99 x 10^9 N m²/C²) * (6.5 x 10^-6 C)² / (0.065 m)²
Simplifying the equation:
F = (8.99 x 10^9 N m²/C²) * (6.5 x 10^-6 C)² / (0.065 m)²
≈ 7.83 N
The weight of the ball can be determined using the formula:
Weight = mass * gravitational acceleration
Since the gravitational force is negligible in this case, we can assume the weight is zero.
Now, using Newton's second law of motion, we can calculate the acceleration of the ball:
F = m * a
Where:
F is the net force acting on the ball (7.83 N)
m is the mass of the ball (200 g = 0.2 kg)
a is the acceleration we want to find
Plugging in the values:
7.83 N = 0.2 kg * a
Solving for a, we get:
a = 7.83 N / 0.2 kg ≈ 39.15 m/s²
Therefore, the acceleration of the released ball is approximately 39.15 m/s².