# Coulomb's Law ౼ Problems and Solutions

1. What is the minimum possible magnitude of the electrostatic force between two point charges separated by a distance of 1m in vacuum? The constant k = 9 × 10⁹ N-m²/C².

The magnitude of electrostatic force of interaction between two point charges is governed by the Coulomb's law. Let there be two point charges q₁ and q₂ separated by a distance d (given in the problem d = 1m).
Then by Coulomb's law, the magnitude of  electric force between them is
The value of the force F depends on k, q₁, q₂ and d. But as k is a constant and 'd' has also been fixed, the only parameters affecting the force are q₁ and q₂. The magnitude of electric force is directly proportional to the product of the magnitudes of the charges. So the force to be minimum, the product of magnitudes of charges has to be minimum. The values of q₁ and q₂ should be chosen to be as less as possible.

We know that the quantity 'charge' is quantized having its quantum ±e = ±1.6 × 10⁻¹⁹ C (same as charge on an electron, in modulus). This is the least possible amount of charge that can exist individually. Thus for minimum value of F, we should take q₁= q₂= e.

2. What is the ratio of electrostatic force to the gravitational force between an electron and a proton kept at a distance r from each other? Given: mₑ=9.11×10⁻³¹ kg, mₚ=1.67×10⁻²⁷ kg, k=9×10⁹ Nm²/C², G=6.67×10⁻¹¹ Nm²/kg².

The electron and the proton will exert electrostatic forces on each other by the virtue of their charges whereas they will also attract each other gravitationally because of their masses. The ratio of the two is what we have been asked for. The expressions for mutual electrostatic forces of attraction and the gravitational attraction between an electron and a proton can be written using the Coulomb's law of electrostatics and the Newton's Law of Gravitation respectively.
Hence the sought ratio is 2.25 × 10³⁹. This means that the electric force is 2.25 × 10³⁹ times stronger in magnitude than the gravitational one. It is noteworthy that the ratio is independent of the separation r between the charged particles.

3. Two point charges +q and +4q are fixed in space at a separation r from each other. At what location on the line joining the two point charges, a test charge can stay in equilibrium? Also comment on the nature of that equilibrium.

The specified system of two point charges is depicted in the following diagram:
Let P be the point at which a test charge can stay in equilibrium. It is apparent that the point P can neither be on the left side of charge q nor on the rightwards of the charge 4q on the line connecting the two point charges as at any point located in these regions the electrostatic forces on the test charge applied by both the point charges would point in same direction (or parallel). The point P must be located somewhere in between the two point charges and on the axis of the system as shown in the above diagram.

Say a test charge qₒ is staying in equilibrium at point P. The equilibrium is possible only because of the equal and opposite electrostatic forces acting on the test charge due to the other two point charges.
Hence the test charge should be placed in between the two point charges at a distance r/3 from the point charge +q. Also for a positive test charge (as usual), the equilibrium would be stable in nature and for a negative charge it will be unstable. One more point to note is that the equilibrium of the test charge is independent of its magnitude and nature whereas the nature of the equilibrium depends on the nature (or sign) of the test charge.

4. Two charged particles, each having a mass of 5g and charge 10⁻⁷C stay in limiting equilibrium on a horizontal table maintaining a separation of 10 cm between them. Find the coefficient of friction between each particle and the table, which is same between each particle and table.

The two identical particles stay in equilibrium because the Coulomb force on each particle is balanced by the frictional force. Also the case is of limiting friction. For each particle we can write,

Therefore the coefficient of limiting friction between each particle and the table is 0.18.