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A foursplate capacitor Consider a capacitor made of four parallel plates with large area \(A\), evenly spaced with small separation \(s .\) The first and third are connected by a wire, as are the second and fourth. What is the capacitance of this system?

Short Answer

Expert verified
The capacitance of the system is \(\frac{\epsilon_0 A}{2s}\).

Step by step solution

01

Model the Capacitor as Two Series Capacitors

The given capacitor can be modeled as two capacitors in series. Each capacitor consists of parallel plates with the first and third plates forming one capacitor, and the second and forth plates forming the other capacitor. The total capacitance is the series combination of the two capacitors. Each individual capacitor has capacitance calculated by \(C= \epsilon_0 \frac{A}{s}\).
02

Apply the Series Capacitance Formula

The formula for capacitors in series is given by \(\frac{1}{C_{\text{total}}} = \frac{1}{C_1} + \frac{1}{C_2} + \dots\). In this case, both capacitors have the same value (\(C\)), so we can simplify the formula to \(\frac{1}{C_{\text{total}}} = \frac{2}{C}\).
03

Solve for Total Capacitance

To determine \(C_{\text{total}}\), we can rewrite the series formula to \(C_{\text{total}}=\frac{C}{2}\). Substituting \(C = \epsilon_0 \frac{A}{s}\) into the formula, we get \(C_{\text{total}}=\frac{\epsilon_0 \frac{A}{s}}{2}\). So the total capacitance of the system is \(\frac{\epsilon_0 A}{2s}\)

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Key Concepts

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

Series Capacitance Formula
When dealing with multiple capacitors connected in series, the total or equivalent capacitance can be found using a specific formula. This is important as it allows the calculation of the overall effect of the capacitors on a circuit without considering each one individually.

To find the series capacitance, the formula is \( \frac{1}{C_{\text{total}}} = \frac{1}{C_1} + \frac{1}{C_2} + \dots \), where \(C_{\text{total}}\) is the equivalent capacitance, and \(C_1, C_2, \dots\) are the capacitances of the individual capacitors in the series. For capacitors in series, the reciprocal of the total capacitance is the sum of the reciprocals of the individual capacitances. This results in a total capacitance that is less than any of the individual capacitances. It is comparable to adding resistors in parallel where the overall resistance decreases.
Equivalent Capacitance
Equivalent capacitance is a way to simplify a complex circuit of capacitors by calculating a single capacitance value that could replace all the capacitors without changing the total capacitance of the circuit. This can significantly simplify the analysis of circuits involving multiple capacitors.

In series combinations, equivalent capacitance is found as mentioned earlier, while in parallel the formula for equivalent capacitance is simply the sum of all individual capacitances. This results in equivalent capacitance being greater than any single capacitance in the parallel circuit. Understanding equivalent capacitance helps in both predicting the behavior of the circuit and in designing circuits to achieve a desired total capacitance.
Electric Field and Capacitance
The relationship between electric field and capacitance is fundamental in understanding how capacitors store energy. A capacitor essentially consists of two conductors separated by an insulating material known as a dielectric. When a potential difference is applied across the conductors, an electric field is established in the dielectric, and this field stores energy.

The capacitance \(C\) can be described by the equation \(C= \epsilon_0 \frac{A}{s}\), where \(\epsilon_0\) is the permittivity of free space, \(A\) is the area of the plates, and \(s\) is the separation between them. The electric field \(E\) in a parallel plate capacitor can be given by \(E = \frac{V}{d}\), where \(V\) is the potential difference and \(d\) is the distance between the plates. For a given voltage, the larger the plate area or the smaller the separation, the greater the capacitance and thus the greater the energy stored.
Parallel Plate Capacitor
A parallel plate capacitor is a common and widely understood type of capacitor used to illustrate the concept of capacitance. It consists of two conductive plates, separated by a small distance with an insulating dielectric material in between. The capacitance of a parallel plate capacitor is directly proportional to the area of the plates and inversely proportional to the distance between them.

An important aspect to grasp is that when voltage is applied, positive charge builds up on one plate while an equal amount of negative charge builds up on the opposite plate, creating an electric field across the dielectric. This field is uniform in the region away from the edges of the plates. One of the simplest expressions for the capacitance of a parallel plate capacitor is \(C= \epsilon \frac{A}{d}\), where \(\epsilon\) is the permittivity of the dielectric material, \(A\) is the area of one plate, and \(d\) is the separation between the plates.

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Most popular questions from this chapter

Moving perpendicular to a wire At the end of Section \(5.9\) we discussed the case where a charge \(q\) moves perpendicular to a wire. Figures \(5.25\) and \(5.26\) show qualtatively why there is a nonzero force on the charge, pointing in the positive \(x\) direction. Carry out the calculation to show that the force at a distance \(\ell\) from the wire equals \(q v l / 2 \pi \epsilon_{0} \ell c^{2}\). That is, use Eq. (5.15) to calculate the force on the charge in its own frame, and then divide by \(\gamma\) to transform back to the lab frame. Notes: You can use the fact that in the charge \(q\) 's frame, the speed of the electrons in the \(x\) direction is \(v_{0} / \gamma\) (this comes from the transverse-velocity-addition formula). Remember that the \(\beta\) in Eq. \((5.15)\) is the velocity of the electrons in the charge's frame, and this velocity has two components. Be careful with the transverse distance involved. There are many things to keep track of in this problem, but the integration itself is easy if you use a computer (or Appendix K).

Maximum power * Show that if a battery of fixed emf \(\mathcal{E}\) and internal resistance \(R_{i}\) is connected to a variable external resistance \(R\), the maximum power is delivered to the external resistor when \(R=R_{i}\).

In the laboratory frame, a proton is at rest at the origin at \(t=0 .\) At that instant a negative pion (charge \(-e\) ) that has been traveling in along the \(x\) axis at a speed of \(0.6 c\) reaches the point \(x=0.01 \mathrm{~cm}\). There are no other charges around. What is the magnitude of the force on the pion? What is the magnitude of the force on the proton? What about Newton's third law? (We're getting a little ahead of ourselves with this last question, but see if you can answer it anyway.)

Resistances in a cube ** A cube has a resistor \(R\) along each edge. Find the equivalent resistance between two nodes that correspond to: (a) diagonally opposite corners of the cube; (b) diagonally opposite corners of a face; (c) adjacent corners. You do not need to solve a number of simultaneous equations; instead use symmetry arguments. Hint: If two vertices are at the same potential, they can be collapsed to one point without changing the equivalent resistance between the two given nodes.

Synchrotron current * In a 6 gigaelectron-volt \(\left(1 \mathrm{GeV}=10^{9} \mathrm{eV}\right)\) electron synchrotron, electrons travel around the machine in an approximately circular path 240 meters long. It is normal to have about \(10^{11}\) electrons circling on this path during a cycle of acceleration. The speed of the electrons is practically that of light. What is the current? We give this very simple problem to emphasize that nothing in our definition of current as rate of transport requires the velocities of the charge carriers to be nonrelativistic and that there is no rule against a given charged particle getting counted many times during a second as part of the current.

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