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(a) Two spheres have radii \(a\) and \(b\) and their centers are a distance \(d\) apart. Show that the capacitance of this system is $$C=\frac{4 \pi \epsilon_{0}}{\frac{1}{a}+\frac{1}{b}-\frac{2}{d}}$$ provided that \(d\) is large compared with \(a\) and \(b\). (Suggestion: Because the spheres are far apart, assume that the potential of each equals the sum of the potentials due to each sphere, and when calculating those potentials assume that \(V=k_{e} Q / r\) applies.) \((\mathrm{b})\) Show that as \(d\) approaches infinity the above result reduces to that of two spherical capacitors in series.

Short Answer

Expert verified
a) The capacitance of the system of two spheres is given by \(C= \frac{4 \pi \epsilon_{0}}{1/a + 1/b - 2/d}\). b) As the distance between the spheres becomes very large, the above result simplifies to the capacitance of two spherical capacitors in series, i.e., \(C= \frac{4 \pi \epsilon_{0}}{1/a + 1/b}\).

Step by step solution

01

Derive the Equation for Capacitance

In the understanding that the potential of each sphere can be assumed as the sum of potentials due to each sphere, and \(V = k_{e} Q / r\) can be applied, the potential of the first sphere due to its own charge can be written as \(V_{1}=k_{e} Q / a\). The potential at the first sphere due to the charge on the second sphere can be assumed as \(V_{2}=k_{e} Q / d\). Then, the resultant potential at the first sphere is given as \(V_{T}=V_{1}+V_{2}=k_{e} Q(1/a + 1/d)\). But, potential \(V = Q/C\), thus we get the capacitance of the first sphere as \(C_{1}= Q/V_{T}= 1 / (4 \pi \epsilon_{0}(1/a + 1/d))\). Similarly, we can find the capacitance of the second sphere \(C_{2}= 1 / (4 \pi \epsilon_{0}(1/b + 1/d))\)
02

Find the Equivalent Capacitance

Noting that the two spheres could be considered in parallel to each other, we can find the total capacitance, \(C\), of the entire system by paralleling the two capacitances as \(1/C = 1/C_{1} + 1/C_{2}\). Substituting the expressions obtained in Step 1 for \(C_{1}\) and \(C_{2}\), we obtain \(C= \frac{4 \pi \epsilon_{0}}{1/a + 1/b - 2/d}\)
03

Find the Limiting Case

Next, check the limit as \(d\) approaches infinity. As the distance between the spheres approaches infinity, the term \(-2/d\) will become insignificant. Thus the expression for \(C\) simplifies to \(C= \frac{4 \pi \epsilon_{0}}{1/a + 1/b}\), which corresponds to the result for the capacitance of two spherical capacitors in series.

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Key Concepts

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

Electric Potential and Capacitance
Electric potential, often denoted as 'V', describes the potential energy per unit charge at a point in an electric field and reflects the work done to move a charge to that point from infinitely far away. A critical aspect of electric potential is its relationship with capacitance. Capacitance indicates a system's ability to store electric charge and is quantified by the amount of charge stored per unit of electric potential difference, described by the equation \( C = \frac{Q}{V} \).

Understanding the relationship between electric potential and capacitance is essential when analyzing electric circuits, such as the one involving two spheres with radii \( a \) and \( b \), separated by a distance \( d \). The total electric potential at any point is the sum of potentials due to individual charges or components in the system. This concept is key to deriving the capacitance for complex configurations like spherical capacitors placed at a distance from each other.
Spherical Capacitors
A spherical capacitor typically consists of two concentric spherical conductors, separated by an insulating material or vacuum. The capacitance of a simple spherical capacitor with a single gap is given by \( C = 4\pi\epsilon_0\frac{ab}{b-a} \), where \( a \) and \( b \) are the radii of the inner and outer spheres, respectively, and \( \epsilon_0 \) is the permittivity of free space.

When dealing with two separate spheres rather than concentric ones, the potential at the surface of each is influenced by the charge on both spheres, not just the charge on itself. This interaction complicates the calculation of capacitance, as we must account for the composite electric potential resulting from both charges and the geometry of the setup.
Series and Parallel Capacitance
Capacitance in an electric circuit can be configured in two primary ways: in series or in parallel. For capacitors connected in series, the total capacitance can be found using the equation \( \frac{1}{C_{\text{total}}} = \sum_{i=1}^{n} \frac{1}{C_i} \), where \( C_i \) represents each individual capacitor's capacitance. In contrast, for capacitors connected in parallel, the total capacitance is the sum of individual capacitances, given by \( C_{\text{total}} = \sum_{i=1}^{n} C_i \).

For two individual spherical capacitors distanced from each other, as in our example, the capacitances are not in series or parallel in the conventional sense. However, when spheres are far apart, we approximate their behavior as if they were in series due to the dominant effect of charge contribution, which leads us to the presented steps in the original solution.
Electric Field and Potential Relationship
The relationship between an electric field (E) and electric potential (V) is both fundamental and profound. The electric field at any point can be defined as the negative gradient of the electric potential, mathematically described by \( E = -abla V \). This relationship states that the electric field points in the direction where the potential decreases most rapidly and its magnitude corresponds to the rate of decrease.

In the context of two separated spheres, each sphere creates its own electric field, contributing to the potential at any given point in space. The final potential at each sphere is a superposition of the potentials due to both spheres. This concept aids in deriving the equation for the capacitance of a system consisting of two spheres, leading to the expression provided in the original exercise.

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Most popular questions from this chapter

The immediate cause of many deaths is ventricular fibrillation, uncoordinated quivering of the heart as opposed to proper beating. An electric shock to the chest can cause momentary paralysis of the heart muscle, after which the heart will sometimes start organized beating again. A defibrillator (Fig. 26.14 ) is a device that applies a strong electric shock to the chest over a time interval of a few milliseconds. The device contains a capacitor of several microfarads, charged to several thousand volts. Electrodes called paddles, about \(8 \mathrm{cm}\) across and coated with conducting paste, are held against the chest on both sides of the heart. Their handles are insulated to prevent injury to the operator, who calls, "Clear!" and pushes a button on one paddle to discharge the capacitor through the patient's chest. Assume that an energy of \(300 \mathrm{J}\) is to be delivered from a \(30.0-\mu \mathrm{F}\) capacitor. To what potential difference must it be charged?

Two identical parallel-plate capacitors, each with capacitance \(C,\) are charged to potential difference \(\Delta V\) and connected in parallel. Then the plate separation in one of the capacitors is doubled. (a) Find the total energy of the system of two capacitors before the plate separation is doubled. (b) Find the potential difference across each capacitor after the plate separation is doubled. (c) Find the total energy of the system after the plate separation is doubled. (d) Reconcile the difference in the answers to parts (a) and (c) with the law of conservation of energy.

A wafer of titanium dioxide \((\kappa=173)\) of area \(1.00 \mathrm{cm}^{2}\) has a thickness of \(0.100 \mathrm{mm}\). Aluminum is evaporated on the parallel faces to form a parallel-plate capacitor. (a) Calculate the capacitance. (b) When the capacitor is charged with a \(12.0-\mathrm{V}\) battery, what is the magnitude of charge delivered to each plate? (c) For the situation in part (b), what are the free and induced surface charge densities? (d) What is the magnitude of the electric field?

An air-filled capacitor consists of two parallel plates, each with an area of \(7.60 \mathrm{cm}^{2},\) separated by a distance of 1.80 mm. A 20.0-V potential difference is applied to these plates. Calculate (a) the electric field between the plates, (b) the surface charge density, (c) the capacitance, and (d) the charge on each plate.

A \(20.0-\mu \mathrm{F}\) spherical capacitor is composed of two concentric metal spheres, one having a radius twice as large as the other. The region between the spheres is a vacuum. Determine the volume of this region.

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