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Question: The electrostatic force on a small sphere of charge \( 0.4\mu C \) due to another small sphere of ch...

The electrostatic force on a small sphere of charge 0.4μC0.4\mu C due to another small sphere of charge −0.8μC- 0.8\mu C in the air is 0.2 N.0.2{\text{ }}N.
(a) What is the distance between the two spheres?
(b) What is the force on the second sphere due to the first?

Explanation

Solution

The electrostatic force between two charges is proportional to the product of the charges of the two spheres. The first sphere exerts an equal amount of force on the second as the seconds exerts on the first.

Formula used: In this solution, we will use the following formula:
Coulomb’s law: The force FF between two charges Q and q: F=kQqr2F = \dfrac{{kQq}}{{{r^2}}} where rr is the distance between these two charges.

Complete step by step answer
We’ve been given that the charge on the first sphere is 0.4μC0.4\mu C and charge on the second sphere is −0.8μC- 0.8\mu C . Now, we know that the force between two charges that are a certain distance rr apart can be written as:
F=kQqr2F = \dfrac{{kQq}}{{{r^2}}}
Substituting the value of Q=0.4 μC=0.4×10−6 CQ = 0.4\,\mu C = 0.4 \times {10^{ - 6}}\,C and q=−0.8 μC=−0.8 ×10−6 Cq = - 0.8\,\mu C = - 0.8\, \times {10^{ - 6}}\,C , F=0.2 N.F = 0.2{\text{ }}N. and where , we get the force as
0.2=14π(8.85×10−12)(0.4×10−6)(−0.8×10−6)r20.2 = \dfrac{{\dfrac{1}{{4\pi \left( {8.85 \times {{10}^{ - 12}}} \right)}}(0.4 \times {{10}^{ - 6}})( - 0.8 \times {{10}^{ - 6}})}}{{{r^2}}}
Simplifying the above equation, we can solve for r2{r^2} as
r2=144×10−4 m{r^2} = 144 \times {10^{ - 4}}\,m
Taking the square root on both sides, we will get the distance between the two charges as:
r=0.12 mr = 0.12\,{\text{m}}
Hence the distance between the two spheres will be r=0.12 mr = 0.12\,{\text{m}} .
According to Newton’s third law, the force exerted by the first sphere on the second will have the same magnitude but have an opposite direction. Hence the force will also be 0.2 N0.2\,N will also be attractive.

Note
The force between two oppositely charged spheres will always be attractive. Here, we have assumed that the distance between the spheres will be larger than the radius of the spheres. This will help us in treating the spheres as a point object and we can then use the force equation that we used.