Answer:When the ball rolls off the edge of the table, it will continue moving forward at 2.0 m/s until it hits the floor.
Explanation:This is what I would say is the answer bc I had to do reasearch on a lot of this for my work this year so if its not im veery sorry
Use Coulomb law: F = k * q1*q2 / (r^2), where k = 9.00 * 10^9 N.m^2/C^2
F = 9.00 * 10^9 N.m^2/C^2 * 2.4*10^-8 C * 1.8*10^-6 C / [0.008m]^2 = 38.88 * 10^ -5 N
F = 39 * 10 -5N
Potential and kinetic energy along with an external force
The original Coulomb force between the charges is:
Fc=(k*Q₁*Q₂)/r², where k is the Coulomb constant and k=9*10⁹ N m² C⁻², Q₁ is the first charge, Q₂ is the second charge and r is the distance between the charges.
The magnitude of the force is independent of the sign of the charge so I can simply say they are both positive.
Q₁ is decreased to Q₁₁=(1/3)*Q₁=Q₁/3 and
Q₂ is decreased to Q₂₂=(1/2)*Q₂=Q₂/2.
New force:
Fc₁=(k*Q₁₁*Q₂₂)r², now we input the decreased values of the charge
Fc₁=(k*{Q₁/3}*{Q₂/2})/r², that is equal to:
Fc₁=(k*(1/3)*(1/2)*Q₁*Q₂)/r²,
Fc₁=(k*(1/6)*Q₁*Q₂)/r²
Fc₁=(1/6)*(k*Q₁*Q₂)/r², and since the original force is: Fc=(k*Q₁*Q₂)/r² we get:
Fc₁=(1/6)*Fc
So the magnitude of the new force Fc₁ with decreased charges is 6 times smaller than the original force Fc.
