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Chapter 11: Problem 1647

Two metal plate form a parallel plate capacitor. The distance between the plates is \(\mathrm{d}\). A metal sheet of thickness \(\mathrm{d} / 2\) and of the same area is introduced between the plates. What is the ratio of the capacitance in the two cases? (A) \(4: 1\) (B) \(3: 1\) (C) \(2: 1\) (D) \(5: 1\)

Short Answer

Expert verified
The ratio of the capacitance in the two cases is (C) \(2:1\).

Step by step solution

01

Understand the capacitance formula for parallel plate capacitor

Capacitance (C) between two plates is given by the formula: \(C = \frac{\epsilon A}{d}\) Where ε is the permittivity of the dielectric material between the plates, A is the area of the plates, and d is the distance between them.
02

Capacitance in the first case (C1)

The first case, the distance between the plates is d. Using the capacitance formula for a parallel plate capacitor: \(C_1 = \frac{\epsilon A}{d}\)
03

Capacitance in the second case (C2)

In the second case, a metal sheet is introduced between the plates and it has a thickness of d/2. The metal sheet divides the capacitor into two capacitors connected in series, each with a thickness of d/2. The new capacitance (C2) is equal to the capacitance of one of these new capacitors multiplied by 2 in series: \(C'_2 = \frac{\epsilon A}{d/2}\) As the two capacitors are in series, their total capacitance (C2) can be found using the formula for capacitors in series: \(\frac{1}{C_2} = \frac{1}{C'_2} + \frac{1}{C'_2}\) Which can be simplified to: \(\frac{1}{C_2} = 2 \cdot \frac{1}{C'_2}\)
04

Determine the ratio of capacitance

We now want the ratio \(\frac{C_1}{C_2}\). Let's plug in the values we found for C1 and C2: \(\frac{C_1}{C_2} = \frac{\frac{\epsilon A}{d}}{\frac{1}{2 \cdot \frac{1}{\frac{\epsilon A}{d/2}}}}\) Simplifying: \(\frac{C_1}{C_2} = \frac{2 \epsilon A d}{d \epsilon A}\) Which yields: \(\frac{C_1}{C_2} = \frac{2}{1}\) Hence, the capacitance ratio in the two cases is 2:1. The correct answer is (C) \(2:1\).

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Most popular questions from this chapter

Two thin wire rings each having a radius \(R\) are placed at a distance \(d\) apart with their axes coinciding. The charges on the two rings are \(+q\) and \(-q\). The potential difference between the centers of the two rings is $\ldots .$ (A) 0 (B) $\left.\left[\mathrm{q} /\left(2 \pi \epsilon_{0}\right)\right]\left[(1 / \mathrm{R})-\left\\{1 / \sqrt{(}^{2}+\mathrm{d}^{2}\right)\right\\}\right]$ (C) $\left[\mathrm{q} /\left(4 \pi \epsilon_{0}\right)\right]\left[(1 / \mathrm{R})-\left\\{1 / \sqrt{\left. \left.\left(\mathrm{R}^{2}+\mathrm{d}^{2}\right)\right\\}\right]}\right.\right.$ (D) \(\left[(q R) /\left(4 \pi \epsilon_{0} d^{2}\right)\right]\)

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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) \(2.1 \times 10^{-9} \mathrm{C}, 4.41 \times 10^{-9} \mathrm{C}\)

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A battery is used to charge a parallel plate capacitor till the potential difference between the plates becomes equal to the electromotive force of the battery. The ratio of the energy stored in the capacitor and work done by the battery will be (A) \((1 / 2)\) (B) \((2 / 1)\) (C) 1 (D) \((1 / 4)\)
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