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Chapter 11: Problem 1610

Two parallel plate air capacitors have their plate areas 100 and $500 \mathrm{~cm}^{2}$ respectively. If they have the same charge and potential and the distance between the plates of the first capacitor is \(0.5 \mathrm{~mm}\), what is the distance between the plates of the second capacitor ? (A) \(0.25 \mathrm{~cm}\) (B) \(0.50 \mathrm{~cm}\) (C) \(0.75 \mathrm{~cm}\) (D) \(1 \mathrm{~cm}\)

Short Answer

Expert verified
The distance between the plates of the second capacitor is \(0.25\,\mathrm{cm}\).

Step by step solution

01

Write the capacitance formula for both capacitors

We will use the formula for the capacitance of a parallel plate capacitor, which is: \(C = \dfrac{ε_0A}{d}\) For the first capacitor, let's denote the capacitance as \(C_1\), the plate area as \(A_1 = 100\,\mathrm{cm}^2\), and the distance between the plates as \(d_1 = 0.5\,\mathrm{mm}\). For the second capacitor, let's denote the capacitance as \(C_2\), the plate area as \(A_2 = 500\,\mathrm{cm}^2\), and the distance between the plates as \(d_2\), which we need to find.
02

Set up the equation to find the distance between the plates of the second capacitor

Since both capacitors have the same charge and potential, their capacitances are equal, i.e., \(C_1 = C_2\). Now, using the capacitance formula, we can write the following equation: \(\dfrac{ε_0A_1}{d_1} = \dfrac{ε_0A_2}{d_2}\) We want to find \(d_2\), so let's rearrange the equation to solve for it: \(d_2 = \dfrac{A_2}{A_1} d_1\)
03

Substitute the given values and solve for the distance

Now, we can substitute the values of \(A_1\), \(A_2\), and \(d_1\) into the equation: \(d_2 = \dfrac{500\,\mathrm{cm}^2}{100\,\mathrm{cm}^2} (0.5\,\mathrm{mm})\) To make the units consistent, let's convert the distance from millimeters to centimeters: \(0.5\,\mathrm{mm} = 0.5\,\mathrm{cm/10} = 0.05\,\mathrm{cm}\) Now, we can plug this value into the equation: \(d_2 = \dfrac{500\,\mathrm{cm}^2}{100\,\mathrm{cm}^2} (0.05\,\mathrm{cm})\)
04

Calculate the distance between the plates of the second capacitor

Now, let's solve for \(d_2\): \(d_2 = 5 (0.05\,\mathrm{cm}) = 0.25\,\mathrm{cm}\) So, the distance between the plates of the second capacitor is \(0.25\,\mathrm{cm}\). Therefore, the correct option is (A) \(0.25\,\mathrm{cm}\).

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Most popular questions from this chapter

64 identical drops of mercury are charged simultaneously to the same potential of 10 volt. Assuming the drops to be spherical, if all the charged drops are made to combine to form one large drop, then its potential will be (A) \(100 \mathrm{~V}\) (B) \(320 \mathrm{~V}\) (C) \(640 \mathrm{~V}\) (D) \(160 \mathrm{~V}\)

A long string with a charge of \(\lambda\) per unit length passes through an imaginary cube of edge \(\ell\). The maximum possible flux of the electric field through the cube will be ....... (A) \(\sqrt{3}\left(\lambda \ell / \in_{0}\right)\) (B) \(\left(\lambda \ell / \in_{0}\right)\) (C) \(\sqrt{2}\left(\lambda \ell / \in_{0}\right)\) (D) \(\left[\left(6 \lambda \ell^{2}\right) / \epsilon_{0}\right]\)

A parallel plate air capacitor has a capacitance \(18 \mu \mathrm{F}\). If the distance between the plates is tripled and a dielectric medium is introduced, the capacitance becomes \(72 \mu \mathrm{F}\). The dielectric constant of the medium is (A) 4 (B) 12 (C) 9 (D) 2

At what angle \(\theta\) a point \(P\) must be located from dipole axis so that the electric field intensity at the point is perpendicular to the dipole axis? (A) \(\tan ^{-1}(1 / \sqrt{2})\) (B) \(\tan ^{-1}(1 / 2)\) (C) \(\tan ^{-1}(2)\) (C) \(\tan ^{-1}(\sqrt{2})\)

A sphere of radius \(1 \mathrm{~cm}\) has potential of \(8000 \mathrm{v}\), then energy density near its surface will be ...... (A) \(64 \times 10^{5}\left(\mathrm{~J} / \mathrm{m}^{3}\right)\) (B) \(2.83\left(\mathrm{~J} / \mathrm{m}^{3}\right)\) (C) \(8 \times 10^{3}\left(\mathrm{~J} / \mathrm{m}^{3}\right)\) (D) \(32\left(\mathrm{~J} / \mathrm{m}^{3}\right)\)
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