/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 61 A capacitor consists of a conduc... [FREE SOLUTION] | 91Ó°ÊÓ

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A capacitor consists of a conducting sphere of radius \(a\) surrounded by a concentric conducting shell of radius \(b .\) Show that its capacitance is \(C=a b / k(b-a)\)

Short Answer

Expert verified
The capacitance \(C\) of a spherical capacitor with an inner sphere of radius \(a\) and a concentric outer shell of radius \(b\) is given by the formula \(C = ab/k(b - a)\), where \(k\) is Coulomb's constant.

Step by step solution

01

Understanding the Problem

The problem describes a capacitor consisting of a conducting sphere of radius \(a\) surrounded by a concentric conducting shell of radius \(b\). The task is to prove that the capacitance \(C\) of this setup is given by the formula \(C=ab/k(b-a)\). In this expression, \(k\) usually denotes the Coulomb constant.
02

Set up the Capitance Equation

The capacitance \(C\) of a capacitor is defined as the ratio of the charge \(Q\) it holds to the voltage \(V\) across it, i.e., \(C = Q/V\). In this scenario, the voltage \(V\) will be the electric potential difference between the inner and outer shells.
03

Compute the Electric Potential

For a spherical charge distribution, the electric potential \(V\) at radius \(r\) from the centre is given by \(V = kQ/r\), where \(Q\) is the charge on the sphere and \(k\) is Coulomb's constant. Applying this to the capacitor setup, the potential difference across the capacitor, from the outer shell to the inner sphere, is \(V = kQ/b - kQ/a = kQ(1/a - 1/b)\).
04

Derive the Capacitance Formula

Substitute the potential \(V\) from step 3 into the capacitance equation from step 2: \(C = Q/V = Q/kQ(1/a - 1/b) = 1/k(1/(1/a - 1/b\))\). Simplifying, we get \(C = ab/k(b - a)\).

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Key Concepts

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

Conducting Sphere
A conducting sphere is a key component in understanding the behavior of spherical capacitors. It consists of a material that allows electric charges to move freely across its surface. These spheres are often used in experiments to create uniform electric fields. Conducting spheres are essential when dealing with capacitors because they help establish a potential difference due to the distribution of charges.
A conducting sphere, when charged, will have its charges evenly distributed on its surface. This distribution occurs because the charges repel each other and move to maintain equilibrium. This property is crucial for deriving the formulas related to capacitance and potential difference in spherical capacitors. It ensures that the electric field within the sphere remains zero, emphasizing the concept that every point on the surface is at the same potential.
Electric Potential Difference
The electric potential difference, often referred to as voltage, is the difference in electric potential between two points in an electric field. For capacitors, it's the potential between the plates or surfaces of the capacitor.
In the case of our spherical capacitor, the electric potential difference is the work done in moving a unit charge from the outer shell (radius \(b\)) to the inner sphere (radius \(a\)). You can think of it as the effort needed to move charges between the shell and the sphere. The potential difference is crucial because it affects how the device stores energy. The larger the potential difference, the more energy the system can store. For spherical capacitors, this potential difference is given by \(V = kQ(1/a - 1/b)\), where \(k\) is Coulomb's constant, and \(Q\) is the charge.
Coulomb's Constant
Coulomb's constant, represented by \(k\), plays a fundamental role in calculating electrical interactions. It is a proportionality factor in electrostatics that appears in Coulomb's Law, describing the electrostatic force between two point charges.
In the spherical capacitor setup, Coulomb's constant is essential for calculating the electric potential and consequently the capacitance. It is expressed in units of \(N \cdot m^2/C^2\) and has a value of approximately \(8.99 \times 10^9\) \(N \cdot m^2/C^2\). This constant allows us to establish the potential difference in capacitors by relating the charge to the electric potential created. Understanding how \(k\) interacts in these equations aids in unraveling the intricacies of electrostatic systems.
Spherical Charge Distribution
Spherical charge distribution refers to the arrangement of electric charge over a spherical surface. In the context of a spherical capacitor, this distribution impacts electric potential and capacitance calculations significantly.
When a charge is placed on a conducting sphere, it spreads evenly across its outer surface due to repulsion between like charges. This uniform distribution simplifies many complex electrostatic calculations, such as the electric potential outside and inside the sphere. The beauty of spherical distribution is that it allows us to treat the sphere as a point charge by evaluating its effects at a distance, thus simplifying the calculation of electric fields and potentials.
  • This uniform distribution leads to a symmetric electric field, which is consistent in all directions away from the sphere's surface.
  • In capacitors, such distribution aids in a straightforward calculation of potential energies and helps derive relations between various physical quantities, just like the capacitance formula for our spherical capacitor system.

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Most popular questions from this chapter

An unknown capacitor \(C\) is connected in series with a \(3.0-\mu \mathrm{F}\) capacitor; this pair is placed in parallel with a 1.0 - \(\mu\) F capacitor, and the entire combination is put in series with a 2.0 - \(\mu\) F capacitor. (a) Make a circuit diagram of this network. (b) When a potential difference of \(100 \mathrm{V}\) is applied across the open ends of the network, the total energy stored in all the capacitors is \(5.8 \mathrm{mJ}\). Find \(C\).

Two capacitors are connected in series and the combination is charged to 100 V. If the voltage across each capacitor is \(50 \mathrm{V}\) how do their capacitances compare?

An uncharged capacitor has parallel plates \(5.0 \mathrm{cm}\) on a side, spaced \(1.2 \mathrm{mm}\) apart. (a) How much work is required to transfer \(7.2 \mu \mathrm{C}\) from one plate to the other? (b) How much work is required to transfer an additional \(7.2 \mu \mathrm{C} ?\)

Three point charges \(+q\) and a fourth, \(-\frac{1}{2} q,\) are assembled to form a square of side \(a\). Find an expression for the electrostatic energy of this charge distribution.

Engineers testing an ultracapacitor (see Application on page 420 ) measure the capacitor's stored energy at different voltages. The table below gives the results. Determine a quantity that, when you plot stored energy against it, should give a straight line. Make your plot, establish a best-fit line, and use its slope to determine the capacitance. $$\begin{array}{|l|l|l|l|l|l|l|l|} \hline \text { Voltage (V) } & 12.2 & 20.1 & 31.8 & 37.9 & 45.7 & 50.2 & 56.0 \\\ \hline \text { Energy (kJ) } & 9.25 & 27.2 & 62.5 & 94 & 139 & 158 & 203 \\ \hline \end{array}$$

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