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Chapter 11: Problem 1527

Two small conducting sphere of equal radius have charges \(+1 \mathrm{c}\) and \(-2 \mathrm{c}\) respectively and placed at a distance \(\mathrm{d}\) from each other experience force \(F_{1}\). If they are brought in contact and separated to the same distance, they experience force \(F_{2}\). The ratio of \(F_{1}\) to \(F_{2}\) is \(\ldots \ldots \ldots \ldots\) (A) \(-8: 1\) (B) \(1: 2\) (C) \(1: 8\) (D) \(-2: 1\)

Short Answer

Expert verified
The ratio of \(F_1\) to \(F_2\) is \(-8:1\).

Step by step solution

01

Calculate force F1

Use Coulomb's Law formula for the force between two charged objects: \(F_1 = k \frac{Q_1 Q_2}{d^2}\) where, k is Coulomb's constant = \(9 \times 10^9 Nm^2C^{-2}\), \(Q_1 = +1c\), \(Q_2 = -2c\), and d is the distance between the spheres. Plug in the values: \(F_1 = 9 \times 10^9 \frac{(+1)(-2)}{d^2}\)
02

Determine redistributed charges

When the spheres come into contact, the total charge becomes, \(Q_{total} = Q_1 + Q_2 = +1c - 2c = -1c\) Since the spheres have equal radius, the charge will be distributed equally between them after touching, so each sphere gets half of the total charge: \(Q_{1(new)} = Q_{2(new)} = -\frac{1}{2}c\)
03

Calculate force F2

Use Coulomb's Law formula again, now with the redistributed charges of the spheres. \(F_2 = k \frac{Q_{1(new)} Q_{2(new)}}{d^2}\) Plug in the values: \(F_2 = 9 \times 10^9 \frac{(-\frac{1}{2})(-\frac{1}{2})}{d^2}\)
04

Calculate the ratio of F1 to F2

Now, we need the ratio of F1 to F2. We can cancel the common factors on both sides, so we get: \(\frac{F_1}{F_2} = \frac{[-2]}{[\frac{1}{4}]}\) Solve it: \(\frac{F_1}{F_2} = -8\)
05

Select the correct option

The ratio of F1 to F2 is -8:1. So, the correct answer is: (A) -8:1

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Most popular questions from this chapter

Two Points \(P\) and \(Q\) are maintained at the Potentials of \(10 \mathrm{v}\) and \(-4 \mathrm{v}\), respectively. The work done in moving 100 electrons from \(\mathrm{P}\) to \(\mathrm{Q}\) is \(\ldots \ldots \ldots\) (A) \(2.24 \times 10^{-16} \mathrm{~J}\) (B) \(-9.60 \times 10^{-17} \mathrm{~J}\) (C) \(-2.24 \times 10^{-16} \mathrm{~J}\) (D) \(9.60 \times 10^{-17} \mathrm{~J}\)

Two identical capacitors have the same capacitance \(\mathrm{C}\). one of them is charged to a potential \(\mathrm{V}_{1}\) and the other to \(\mathrm{V}_{2}\). The negative ends of the capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is (A) \((1 / 4) \mathrm{C}\left(\mathrm{V}_{1}^{2}-\mathrm{V}_{2}^{2}\right)\) (B) \((1 / 4) \mathrm{C}\left(\mathrm{V}_{1}^{2}+\mathrm{V}_{2}^{2}\right)\) (C) \((1 / 4) \mathrm{C}\left(\mathrm{V}_{1}-\mathrm{V}_{2}\right)^{2}\) (D) \((1 / 4) \mathrm{C}\left(\mathrm{V}_{1}+\mathrm{V}_{2}\right)^{2}\)

A parallel plate capacitor of capacitance \(5 \mu \mathrm{F}\) and plate separation \(6 \mathrm{~cm}\) is connected to a \(1 \mathrm{~V}\) battery and charged. A dielectric of dielectric constant 4 and thickness \(4 \mathrm{~cm}\) is introduced between the plates of the capacitor. The additional charge that flows into the capacitor from the battery is (A) \(2 \mu \mathrm{c}\) (B) \(5 \mu \mathrm{c}\) (C) \(3 \mu \mathrm{c}\) (D) \(10 \mu \mathrm{c}\)

Two charged spheres of radii \(R_{1}\) and \(R_{2}\) having equal surface charge density. The ratio of their potential is ..... (A) \(\left(\mathrm{R}_{2} / \mathrm{R}_{1}\right)\) (B) \(\left(\mathrm{R}_{2} / \mathrm{R}_{1}\right)^{2}\) (C) \(\left(\mathrm{R}_{1} / \mathrm{R}_{2}\right)^{2}\) (D) \(\left(\mathrm{R}_{1} / \mathrm{R}_{2}\right)\)

Let $\mathrm{P}(\mathrm{r})\left[\mathrm{Q} /\left(\pi \mathrm{R}^{4}\right)\right] \mathrm{r}$ be the charge density distribution for a solid sphere of radius \(\mathrm{R}\) and total charge \(\mathrm{Q}\). For a point ' \(\mathrm{P}\) ' inside the sphere at distance \(\mathrm{r}_{1}\) from the centre of the sphere the magnitude of electric field is (A) $\left[\mathrm{Q} /\left(4 \pi \epsilon_{0} \mathrm{r}_{1}^{2}\right)\right]$ (B) $\left[\left(\mathrm{Qr}_{1}^{2}\right) /\left(4 \pi \in{ }_{0} \mathrm{R}^{4}\right)\right]$ (C) $\left[\left(\mathrm{Qr}_{1}^{2}\right) /\left(3 \pi \epsilon_{0} \mathrm{R}^{4}\right)\right]$
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