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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

If a charged spherical conductor of radius \(10 \mathrm{~cm}\) has potential \(\mathrm{v}\) at a point distant \(5 \mathrm{~cm}\) from its centre, then the potential at a point distant \(15 \mathrm{~cm}\) from the centre will be $\ldots . .$ (A) \((1 / 3) \mathrm{V}\) (B) \((3 / 2) \mathrm{V}\) (C) \(3 \mathrm{~V}\) (D) \((2 / 3) \mathrm{V}\)

Three identical spheres each having a charge \(\mathrm{q}\) and radius \(R\), are kept in such a way that each touches the other two spheres. The magnitude of the electric force on any sphere due to other two is \(\ldots \ldots \ldots\) (A) $(\mathrm{R} / 2)\left[1 /\left(4 \pi \epsilon_{0}\right)\right](\sqrt{5} / 4)(\mathrm{q} / \mathrm{R})^{2}$ (B) $\left[1 /\left(8 \pi \epsilon_{0}\right)\right](\sqrt{2} / 3)(\mathrm{q} / \mathrm{R})^{2}$ (C) $\left[1 /\left(4 \pi \epsilon_{0}\right)\right](\sqrt{3} / 4)(\mathrm{q} / \mathrm{R})^{2}$ (D) $-\left[1 /\left(8 \pi \epsilon_{0}\right)\right](\sqrt{3} / 2)(\mathrm{q} / \mathrm{R})^{2}$

Point charges \(4 \mu \mathrm{c}\) and \(2 \mu \mathrm{c}\) are placed at the vertices \(\mathrm{P}\) and Q of a right angle triangle \(P Q R\) respectively. \(Q\) is the right angle, \(\mathrm{PR}=2 \times 10^{-2} \mathrm{~m}\) and \(\mathrm{QR}=10^{-2} \mathrm{~m}\). The magnitude and direction of the resultant electric field at \(\mathrm{R}\) is \(\ldots \ldots\) (A) \(4.28 \times 10^{9} \mathrm{NC}^{-1}, 45^{\circ}\) (B) \(2.38 \times 10^{8} \mathrm{NC}^{-1}, 40.9^{\circ}\) (C) \(1.73 \times 10^{4} \mathrm{NC}^{-1}, 34.7^{\circ}\) (D) \(4.9 \times 10^{10} \mathrm{NC}^{-1}, 34.7^{\circ}\)

A \(5 \mu \mathrm{F}\) capacitor is charged by a \(220 \mathrm{v}\) supply. It is then disconnected from the supply and is connected to another uncharged $2.5 \mu \mathrm{F}$ capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation ? (A) \(0.02 \mathrm{~J}\) (B) \(0.121 \mathrm{~J}\) (C) \(0.04 \mathrm{~J}\) (D) \(0.081 \mathrm{~J}\)

Equal charges \(q\) are placed at the vertices \(A\) and \(B\) of an equilateral triangle \(\mathrm{ABC}\) of side \(\mathrm{a}\). The magnitude of electric field at the point \(c\) is \(\ldots \ldots \ldots\) (A) \(\left(\mathrm{Kq} / \mathrm{a}^{2}\right)\) (B) \(\left.(\sqrt{3} \mathrm{Kq}) / \mathrm{a}^{2}\right)\) (C) \(\left.(\sqrt{2} \mathrm{Kq}) / \mathrm{a}^{2}\right)\) (D) $\left[\mathrm{q} /\left(2 \pi \mathrm{t} \varepsilon_{0} \mathrm{a}^{2}\right)\right]$
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