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Distribution of charge on a capacitor Consider a parallel-plate capacitor with different magnitudes of charge on the two plates. Let the charges be \(Q_{1}\) and \(Q_{2}\) (which we normally set equal to \(Q\) and \(-Q\) ). Find the four amounts of charge on the inner and outer surfaces of the two plates.

Short Answer

Expert verified
If the charges on the plates of a capacitor are \(Q_{1}\) and \(Q_{2}\), then the charges on the inner and outer surfaces of both plates will respectively be \(Q_{1}/2\) and \(Q_{2}/2\).

Step by step solution

01

Determine the Charge on the Inner Surfaces

As with any parallel plate capacitor, on the inner surfaces, the charges will distribute such that they neutralize each other. This is due to the fact that insides of conductors have zero net electric field. Therefore, if \(Q_{1}\) is the charge on the first plate and \(Q_{2}\) is the charge on the second plate, then the charges on the inner surfaces of the plates will be \(Q_{1}/2\) and \(Q_{2}/2\) respectively.
02

Determine the Charge on the Outer Surfaces

The net charge on each plate must equal the sum of the charges on the inner and outer surfaces of that plate. It means the charge on the outer surface of the first plate is \(Q_{1} - Q_{1}/2 = Q_{1}/2\), and the charge on the outer surface of the second plate is \(Q_{2} - Q_{2}/2 = Q_{2}/2\).
03

Consolidate Results

To consolidate everything, the amount of charge on the inner and outer surfaces of each plate is, therefore, \(Q_{1}/2\) and \(Q_{2}/2\) respectively.

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Key Concepts

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

Capacitor Charge Distribution
Understanding the distribution of charge in a parallel-plate capacitor is essential when analyzing its behavior in an electrical circuit. On a basic level, capacitors are designed to hold and store electrical charge. When two plates of a capacitor are brought close together without touching, a charge is deposited on one plate, and a corresponding negative charge is induced on the second plate.

When we consider a capacitor where the plates have different magnitudes of charge, say, with quantities labeled as \(Q_{1}\) and \(Q_{2}\), it's important to recognize that the charges distribute in a specific way. The distribution occurs such that the electric field within the conductor remains zero; this is a key principle in electrostatics. The charges will arrange themselves to neutralize the electric field inside, resulting in equal but opposite charges on the facing surfaces. As stated in the step-by-step solution, the charges on the inner surfaces of the plates would be \(Q_{1}/2\) and \(Q_{2}/2\) respectively.

It might be tempting to think that these charges remain equally distributed at all times, but when connected to a circuit, the distribution can change depending on various factors like voltage applied and the presence of dielectric materials. As you work through more complex problems involving capacitors, you'll need to apply fundamentals of charge distribution to predict and explain these behaviors.
Electric Field in Conductors
The concept of the electric field in conductors underlies much of what we know about capacitor functionality. Within a conductor in electrostatic equilibrium, such as the plates of a capacitor, the electric field is always zero. This principle is crucial because it's the mechanism that allows the charges to distribute across the surfaces in a particular way.

The reasoning behind this is that if there were an electric field present in the conductor, the free electrons would move in response to it. They would continue to move until they have canceled the field within the material. Thus, in the case of a capacitor, this results in charges on the inner surfaces that exactly counter-balance one another, ensuring zero net electric field within the conductor. Any external electric field applied across a capacitor will affect the charge distribution across the plates. This response by the charges to balance out the electric field is key to understanding why capacitors work and how they store energy.
Net Charge on Capacitors
When dealing with capacitors, particularly parallel-plate capacitors, knowing about the net charge is fundamental. The net charge on an isolated capacitor plate equals the algebraic sum of the charges on both the inner and outer surfaces of that plate. In our given example, when a capacitor has charges \(Q_{1}\) and \(Q_{2}\) on its two plates, the net charge is determined by balancing the charges on their respective inner and outer surfaces, as indicated in the solution steps.

It's crucial to understand that in a typical parallel-plate capacitor, if the charges are equal and opposite, the net charge of the entire capacitor is zero. However, the presence of individual charges on each plate is what creates the electric field and potential difference. Therefore, even if the net charge is zero, the capacitor is capable of storing energy in the form of an electric field between the plates.

The understanding of net charge becomes essential when looking at how capacitors function in circuits. The ability of a capacitor to influence voltage and current depends on the net charge it holds over a period, this ability has numerous practical applications in electronics, such as timing, filtering, and energy storage.

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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).

Capacitance of raindrops \(N\) charged raindrops with radius \(a\) all have the same potential. Assume that they are far enough apart so that the charge distribution on each isn't affected by the others (that is, it is spherically symmetric). What is the total capacitance of this system? How does this capacitance compare with the capacitance in the case where the drops are combined into one big drop?

If a point charge is located outside a hollow conducting shell, there is an electric field outside, but no electric field inside. On the other hand, if a point charge is located inside a hollow conducting shell, there is an electric field both inside and outside (although the external field would be zero in the special case where the shell happened to have charge exactly equal and opposite to the point charge). The situation is therefore not symmetric with respect to inside and

Transatlantic telegraphic cable *? The first telegraphic messages crossed the Atlantic in 1858 , by a cable \(3000 \mathrm{~km}\) long laid between Newfoundland and Ireland. The conductor in this cable consisted of seven copper wires, each of diameter \(0.73 \mathrm{~mm}\), bundled together and surrounded by an insulating sheath. (a) Calculate the resistance of the conductor. Use \(3.10^{-8}\) ohmmeter for the resistivity of the copper, which was of somewhat dubious purity. (b) A return path for the current was provided by the ocean itself. Given that the resistivity of seawater is about \(0.25\) ohm-meter, see if you can show that the resistance of the ocean return would have been much smaller than that of the cable. (Assume that the electrodes immersed in the water were spheres with radius, say, \(10 \mathrm{~cm}\).)

Attenuator chain Some important kinds of networks are infinite in extent. Figure \(4.49\) shows a chain of series and parallel resistors stretching off endlessly to the right. The line at the bottom is the resistanceless return wire for all of them. This is sometimes called an attenuator chain, or a ladder network. The problem is to find the "input resistance," that is, the equivalent resistance between terminals \(A\) and \(B\). Our interest in this problem mainly concerns the method of solution, which takes an odd twist and which can be used in other places in physics where we have an iteration of identical devices (even an infinite chain of lenses, in optics). The point is that the input resistance (which we do not yet know - call it \(R\) ) will not be changed by adding a new set of resistors to the front end of the chain to make it one unit longer. But now, adding this section, we see that this new input resistance is just \(R_{1}\) in series with the parallel combination of \(R_{2}\) and \(R\) Use this strategy to determine \(R\). Show that, if voltage \(V_{0}\) is applied at the input to such a chain, the voltage at successive nodes decreases in a geometric series. What should the ratio of the resistors be so that the ladder is an attenuator that halves the voltage at every step? Obviously a truly infinite ladder would not be practical. Can you suggest a way to terminate it after a few sections without introducing any error in its attenuation?

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