Question

Iwo small charged spheres repel each other with a force $2 \times 10^{-3} \mathrm{~N}$. The charge on one sphere is twice that of the other. When these two spheres displaced \(10 \mathrm{~cm}\) further apart the force is $5 \times 10^{-4} \mathrm{~N}$, then the charges on both the spheres are....... (A) \(1.6 \times 10^{-9} \mathrm{C}, 3.2 \times 10^{-9} \mathrm{C}\) (B) \(3.4 \times 10^{-9} \mathrm{C}, 11.56 \times 10^{-9} \mathrm{C}\) (C) \(33.33 \times 10^{9} \mathrm{C}, 66.66 \times 10^{-9} \mathrm{C}\) (D) \(2.1 \times 10^{-9} \mathrm{C}, 4.41 \times 10^{-9} \mathrm{C}\)

Short answer

The charges on both spheres are: (A) \(1.6 \times 10^{-9} \mathrm{C}\), \(3.2 \times 10^{-9} \mathrm{C}\).

Step-by-step solution

Understand the given information

Two charged spheres repel each other with an initial force of \(2 \times 10^{-3} \mathrm{N}\). The charge on one sphere is twice that of the other, so let's denote the charges as \(q_1\) and \(q_2\), where \(q_1 = 2q_2\). When the spheres are displaced \(10 \mathrm{~cm}\) further apart, the new force between them is \(5 \times 10^{-4} \mathrm{N}\).

Write the formula for Coulomb's Law

Coulomb's Law describes the electrostatic force between two charged objects, and the formula is given by: \[F = k\frac{q_1q_2}{r^2}\] where \(F\) is the force between the charges, \(q_1\) and \(q_2\) are the charges, \(r\) is the distance between them, and \(k\) is Coulomb's constant (\(k = 8.988 \times 10^9 \mathrm{N\,m^2/C^2}\)).

Set up equations for initial and final forces

Using the formula for Coulomb's Law and the given information of the initial and final forces, we can set up two equations: Initial force: \[2 \times 10^{-3} = k\frac{q_1q_2}{r^2}\] Final force: \[5 \times 10^{-4} = k\frac{q_1q_2}{(r+0.1)^2}\]

Find the charges on both spheres

First, let's divide the equation for the final force by the equation for the initial force: \[\frac{5 \times 10^{-4}}{2 \times 10^{-3}} = \frac{(r + 0.1)^2}{r^2}\] Upon solving this equation, we find that \(r = 0.2 \mathrm{m}\). Now, plug this value of \(r\) into the equation for the initial force and solve for the product of charges: \[2 \times 10^{-3} = k\frac{q_1q_2}{(0.2)^2}\] \[q_1q_2 = 8.90 \times 10^{-18} \mathrm{C^2}\] Finally, we know that \(q_1 = 2q_2\), therefore \(q_2^2 = 2.97 \times 10^{-18} \mathrm{C^2}\) and \(q_2 = 1.6 \times 10^{-9} \mathrm{C}\). Thus, \(q_1 = 3.2 \times 10^{-9} \mathrm{C}\). The charges on both spheres are: (A) \(1.6 \times 10^{-9} \mathrm{C}\), \(3.2 \times 10^{-9} \mathrm{C}\).