Question

You have two identical metal spheres, one with a -2.0-µC charge and one with a 20.0-µC charge, located 5 cm apart. They are brought into contact with each other, which distributes the charge evenly over the available surface area, and then moved back to their original positions. Determine the ratio of the magnitude of the new electric force between them to the original electric force between them.

Answer #1

Electrostatic force is given by:

F = k*q1*q2/r^2

Initial electric force between both metal spheres will be:

q1 = -2.0 uC & q2 = 20.0 uC, r1 = distance between both spheres = 5 cm = 0.05 m

So,

|F1| = 9*10^9*2.0*10^-6*20.0*10^-6/0.05^2

|F1| = 144 N

Now when both spheres are brought into contact and then separated, then new charge on both spheres will be:

q1' = q2' = (q1 + q2)/2 = (-2.0 + 20.0)/2 = 9.0 uC

So Now new electric force will be:

|F2| = k*q1'*q2'/r1^2

|F2| = 9*10^9*9.0*10^-6*9.0*10^-6/0.05^2

|F2| = 291.6 N

Now ratio between magnitudes of new and original electric force will be:

F2/F1 = 291.6/144

**F2/F1 = 2.025**

**Let me know if you've any query.**

5. Two identical metal spheres are placed 15.0 cm apart. A
charge of 6.00 µC is placed on one sphere while a charge of −2.00
µC is placed upon the other. What is the force on each sphere? If
the two spheres are brought together and touched and then separated
to their original separation, what will be the force on each
sphere?
Answer: F12 =4.80 N
attractive q = 2.00
μC
F12 = 1.60 N repulsive
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Magnitude
NDirection
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