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Question

The electrostatic force on a small sphere of charge 0.4 μC due to another small sphere of charge –0.8 μC in air is 0.2 N.
What is the distance between the two spheres?
What is the force on the second sphere due to the first?

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Solution

Let us consider two charges q1 and q2. According to the question, q1=0.4μC, q2=0.8μC



Let the distance between two charges be r. Force on the charge q1(0.4μC) due to another charge q2(0.8μC) is
F=0.2N
We have to find the value of r. Using Coulomb's law, the force between two charges is
F=14πε0.q1q2r2
Putting the values of F, 14πε0, q1 and q2, we get
0.2=9×109×0.4×106×0.8×106r2r2=16×9×104r=4×3×102r=12×102mr=12cm
Here, the charge q2 is negative in nature and q1 is positive in nature. So, the force between q1 and q2 will be attractive in nature as unlike charges attract each other.

The force attraction on the second sphere due to the first sphere is same as the force between the two charges i.e., 0.2N. Electrostatic force between two charges obeys the Newton's third law of action-reaction law.


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