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Chapter 11: Problem 1538

Two identical balls having like charges and placed at a certain distance apart repel each other with a certain force. They are brought in contact and then moved apart to a distance equal to half their initial separation. The force of repulsion between them increases \(4.5\) times in comparison with the initial value. The ratio of the initial charges of the balls is ....... (A) \(4: 1\) (B) \(6: 1\) (C) \(3: 1\) (D) \(2: 1\)

Short Answer

Expert verified
The ratio of the initial charges of the balls is \(\boxed{\text{(C)}\; 3:1}\).

Step by step solution

01

Write down the Coulomb's law formula

Coulomb's law states that the force between two point charges is given by: \[F = k\frac{q_1q_2}{r^2}\] where, \(F\) is the force between charges, \(q_1\) and \(q_2\) are the magnitudes of charges, \(r\) is the separation between charges, and \(k\) is Coulomb's constant.
02

Set up two equations for the two cases

Let the initial charges be \(q_1\) and \(q_2\). The initial force and separation can be represented as \(F_1\) and \(r_1\). Using the Coulomb's law, \[F_1 = k\frac{q_1q_2}{(r_1)^2}\] When the balls are brought in contact, the charges become \(\frac{q_1+q_2}{2}\) for both balls. Then, they are moved apart to a distance of \(\frac{r_1}{2}\). The force between them increases 4.5 times to become \(4.5F_1\). \[4.5F_1 = k\frac{\left(\frac{q_1+q_2}{2}\right)^2}{\left(\frac{r_1}{2}\right)^2}\]
03

Manipulate the equations to eliminate variables

From the first equation, we can write: \[k \frac{q_1q_2}{(r_1)^2} = F_1 \] Now, substituting this expression for \(F_1\) in the second equation: \[4.5 k \frac{q_1q_2}{(r_1)^2} = k\frac{\left(\frac{q_1+q_2}{2}\right)^2}{\left(\frac{r_1}{2}\right)^2}\] Cancelling out \(k\) and \((r_1)^2\) from both sides, and simplifying, we get: \[4.5(q_1q_2) = \left(\frac{q_1 + q_2}{2}\right)^2\] Expanding and rearranging: \[\frac{q_1^2}{2} + \frac{q_2^2}{2} = 3.5q_1q_2\]
04

Solve for the ratio of the charges, \(q_1\) and \(q_2\)

Divide the equation by \(q_1q_2\): \[\frac{q_1}{2q_2} + \frac{q_2}{2q_1} = 3.5\] Let \(\frac{q_1}{q_2} = x\), then the equation becomes a quadratic equation in \(x\): \[\frac{x}{2} + \frac{1}{2x} = 3.5\] Solving the quadratic equation, we get \(x = \frac{q_1}{q_2} = 3\), which means the ratio of the initial charges is \(3: 1\). Therefore, the answer is \(\boxed{\text{(C)}\; 3:1}\).

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Most popular questions from this chapter

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