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Chapter 17: Problem 107

The capacitance of a spherical condensers is \(1 \mu \mathrm{F}\). If the spacing between two spheres is \(1 \mathrm{~mm}\), the radius of the outer sphere is (a) \(3 \mathrm{~m}\) (b) \(7 \mathrm{~m}\) (c) \(8 \mathrm{~m}\) (d) \(9 \mathrm{~m}\)

Short Answer

Expert verified
The radius of the outer sphere is approximately 9 m, option (d).

Step by step solution

01

Understand the Formula for Capacitance

The capacitance of a spherical capacitor is given by the formula:\[ C = 4\pi\varepsilon_0 \frac{ab}{b-a} \]where \(a\) is the radius of the inner sphere, \(b\) is the radius of the outer sphere, and \(\varepsilon_0\) is the permittivity of free space, approximately \(8.854 \times 10^{-12} \mathrm{~F/m}\). The distance between the spheres is \(b-a = 1 \mathrm{~mm} = 0.001 \mathrm{~m}\). Given that the capacitance \(C\) is \(1 \mu\mathrm{F} = 1 \times 10^{-6} \mathrm{~F}\).
02

Substitute Known Values into the Formula

Substitute the known values into the formula:\[ 1 \times 10^{-6} = 4\pi (8.854 \times 10^{-12}) \frac{a(a+0.001)}{0.001} \]which simplifies to:\[ 1 \times 10^{-6} = 4\pi (8.854 \times 10^{-12}) a \left(a + 0.001\right) \]
03

Rearrange and Solve for a

Rearrange the equation to find the radius \(a\) of the inner sphere:\[ a(a+0.001) = \frac{1 \times 10^{-6} \times 0.001}{4\pi \times 8.854 \times 10^{-12}} \]Calculate:\[ a(a+0.001) \approx \frac{10^{-9}}{1.11 \times 10^{-10}} \approx 9 \mathrm{~m} \]
04

Calculate Radius of Outer Sphere

Since \(a = 9 \mathrm{~m}\) is the radius of the inner sphere, and given \(b-a = 0.001 \mathrm{~m}\), the radius \(b\) of the outer sphere is:\[ b = a + 0.001 = 9 + 0.001 \approx 9 \mathrm{~m} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Spherical Capacitors
Spherical capacitors are a fascinating electrical component, crucial in various electronic applications. They are made up of two concentric spherical conductors. The inner sphere acts as one electrode, while the outer sphere is the other.
The unique structure of spherical capacitors provides certain advantages, especially concerning their electrical characteristics. Their symmetry contributes to predictable behavior under different electrical conditions.
Applications include coupling and decoupling capacitors in circuits, as well as energy storage components in devices requiring specific capacitance properties. When dealing with spherical capacitors, it helps to visualize them as two hollow metal balls, one encapsulating the other. Remember, the space between the two spheres is a key feature, as this is where the electric field is concentrated.
Capacitance Formula
The capacitance formula for spherical capacitors is an essential tool for predicting their electrical properties. The formula is expressed as:
  • \[ C = 4\pi\varepsilon_0 \frac{ab}{b-a} \]
Here, \( C \) represents capacitance, \( a \) and \( b \) are the radii of the inner and outer spheres, respectively, and \( \varepsilon_0 \) denotes the permittivity of free space.
This formula lets us determine how much electric charge a spherical capacitor can store at a given potential difference.
A crucial aspect of this formula is the term \( \frac{ab}{b-a} \), which considers both the size of the spheres and the spacing between them. This spacing is significant, as small changes can greatly affect the capacitance value. Therefore, paying attention to these parameters is vital when designing or analyzing spherical capacitors.
Permittivity of Free Space
The permittivity of free space, often denoted as \( \varepsilon_0 \), is a fundamental constant in electromagnetism. Its approximate value is \( 8.854 \times 10^{-12} \mathrm{~F/m} \) (farads per meter).
This constant plays a critical role in determining the capacitance of spherical capacitors, as it appears directly in the capacitance formula.
Permittivity describes how an electric field affects and is affected by a dielectric medium, in this case, free space. It essentially measures the ability of the medium to allow electric field lines to pass through it.
Understanding permittivity is crucial for anyone studying capacitors or advanced electronics.
In the context of spherical capacitors, permittivity governs the relationship between the stored charge, voltage, and the physical configuration of the capacitors. As a result, it's essential for both theoretical and practical applications in designing electronic components.

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Most popular questions from this chapter

A spherical charged conductor has \(\sigma\) as the surface density of charge. The electric field on its surface is \(E\). If the radius of the sphere is doubled, keeping the surface density of the charge unchanged, what will be the electric field on the surface of the new sphere? (a) \(\frac{E}{4}\) (b) \(\frac{E}{2}\) (c) \(E\) (d) \(2 \bar{E}\)

A point \(Q\) lies on the perpendicular bisector of an electrical dipole of dipole moment \(p\). If the distance of \(Q\) from the dipole is \(r\) (much larger than the size of the dipole), then the electric field intensity \(E\) at \(Q\) is proportional to (a) \(r^{-2}\) (b) \(r^{-4}\) (c) \(r^{-1}\) \((d) r^{-3}\)

Two free protons are separated by a distance of \(1 \AA\). If one proton is kept at least distance and the other is released, the kinetic energy of second proton when it is at infinite separation is (a) \(23.0 \times 10^{-19} \mathrm{~J}\) (b) \(11.5 \times 10^{-19} \mathrm{~J}\) (c) \(2.3 \times 10^{-19} \mathrm{~J}\) (d) zero

A glass rod rubbed with silk is used to charged a gold leaf electroscope and the leaves are observed to diverse. The electroscope thin, charged is exposed to X-rays for short period. Then, (a) the leaves will diverge further (b) the leaves will melt (c) the leaves will not be affected (d) None of the above

A parallel plate capacitor is made of two dielectric blocks in series. One of the blocks has thickness \(d_{1}\) and dielectric constant \(K_{1}\) and the other has thickness \(d_{2}\) and dielectric constant \(K_{2}\) as shown in figure. This arrangement can be thought as a dielectric slab of thickness \(d\left(=d_{1}+d_{2}\right)\) and effective dielectric constant \(K .\) The \(K\) is \(\quad\) [NCERT Exemplar] (a) \(\frac{K_{1} d_{1}+K_{2} d_{2}}{d_{1}+d_{2}}\) (b) \(\frac{K_{1} d_{1}+K_{2} d_{2}}{K_{1}+K_{2}}\) (c) \(\frac{K_{1} K_{2}\left(d_{1}+d_{2}\right)}{\left(K_{1} d_{1}+K_{2} d_{2}\right)}\) (d) \(\frac{2 K_{1} K_{2}}{K_{1}+K_{2}}\)
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