4. Two charged metal spheres Separated by 20 cm. attract each other with a 12 N resulting force when the charge on each sphere is reduced by half.(7.2) (a) 3 N (b) 6N (c) 24 N (d) 48 N

Question

4. Two charged metal spheres Separated by 20 cm. attract each other with a 12 N resulting force when the charge on each sphere is reduced by half.(7.2) (a) 3 N (b) 6N (c) 24 N (d) 48 N

Answer 1

The force between two charged objects is given by Coulomb's Law:F = k * (q1 * q2) / r^2Where:* F is the force* k is Coulomb's constant* q1 and q2 are the charges* r is the distance between the chargesIn this problem, the distance remains constant, so we can focus on the relationship between the force and the charges. If the charge on *each* sphere is reduced by half, the new charges become q1/2 and q2/2. Let's call the new force F'.F' = k * (q1/2 * q2/2) / r^2F' = k * (q1 * q2) / (4 * r^2)F' = (1/4) * k * (q1 * q2) / r^2Notice that k * (q1 * q2) / r^2 is the original force, F. Therefore:F' = (1/4) * FF' = (1/4) * 12 NF' = 3 NSo the answer is (a) 3 N.

Question

4. Two charged metal spheres Separated by 20 cm. attract each other with a 12 N resulting force when the charge on each sphere is reduced by half.(7.2) (a) 3 N (b) 6N (c) 24 N (d) 48 N

Solution

Answer