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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

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) $\left[\left(\mathrm{Qr}_{1}^{2}\right) /\left(3 \pi \epsilon_{0} \mathrm{R}^{4}\right)\right]$

Two identical metal plates are given positive charges \(\mathrm{Q}_{1}\) and \(\mathrm{Q}_{2}\left(<\mathrm{Q}_{1}\right)\) respectively. If they are now brought close to gather to form a parallel plate capacitor with capacitance \(\mathrm{c}\), the potential difference between them is (A) \(\left[\left(Q_{1}+Q_{2}\right) /(2 c)\right]\) (B) \(\left[\left(\mathrm{Q}_{1}+\mathrm{Q}_{2}\right) / \mathrm{c}\right]\) (C) \(\left[\left(\mathrm{Q}_{1}-\mathrm{Q}_{2}\right) /(2 \mathrm{c})\right]\) (D) \(\left[\left(\mathrm{Q}_{1}-\mathrm{Q}_{2}\right) / \mathrm{c}\right]\)

Large number of capacitors of rating $10 \mu \mathrm{F} / 200 \mathrm{~V} \mathrm{~V}$ are available. The minimum number of capacitors required to design a \(10 \mu \mathrm{F} / 700 \mathrm{~V}\) capacitor is (A) 16 (B) 8 (C) 4 (D) 7

Electric charges of \(+10 \mu \mathrm{c}, 5 \mu \mathrm{c},-3 \mu \mathrm{c}\) and \(8 \mu \mathrm{c}\) are placed at the corners of a square of side $\sqrt{2 m}\( the potential at the centre of the square is \)\ldots \ldots$ (A) \(1.8 \mathrm{~V}\) (B) \(1.8 \times 10^{5} \mathrm{~V}\) (C) \(1.8 \times 10^{6} \mathrm{~V}\) (D) \(1.8 \times 10^{4} \mathrm{~V}\)

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) \((1 / 4) \mathrm{C}\left(\mathrm{V}_{1}+\mathrm{V}_{2}\right)^{2}\)
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