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A parallel-plate capacitor with plate separation \(d\) has the space between the plates filled with two slabs of dielectric, one with constant \(K_{1}\) and the other with constant \(K_{2, \text { and each }}\) having thickness \(d / 2\) . (a) Show that the capacitance is given by \(C=\frac{2 \epsilon_{o} A}{d}\left(\frac{K_{1} K_{2}}{K_{1}+K_{2}}\right) \cdot(\)Hint: Can you think of this combination as two capacitors in series? (b) To see if your answer is reasonable, check it in the following cases: (i) There is only one dielectric, with constant \(K,\) and it completely fills the space between the plates. (ii) The plates have nothing but air, which we can treat as vacuum, between them.

Short Answer

Expert verified
The capacitance is \( C=\frac{2 \epsilon_{o} A}{d}\left(\frac{K_{1} K_{2}}{K_{1}+K_{2}}\right) \). For single dielectric \( C=\frac{\epsilon_{o} K A}{d} \) and for air \( C=\frac{\epsilon_{o} A}{d} \).

Step by step solution

01

Identify the Problem and System

We have a parallel-plate capacitor filled with two dielectrics of different constants, each occupying half the distance between the plates. We need to calculate the capacitance and consider this system as two capacitors in series, each with its own dielectric constant, forming capacitor sections: one with dielectric constant \( K_1 \) and the other with \( K_2 \).
02

Consider the Capacitor Sections in Series

When capacitors are in series, the reciprocal of the total capacitance \( C_{total} \) is the sum of the reciprocals of individual capacitances:\[ \frac{1}{C_{total}} = \frac{1}{C_1} + \frac{1}{C_2} \]Here \( C_1 \) and \( C_2 \) will be the capacitances of the sections with dielectrics \( K_1 \) and \( K_2 \) respectively.
03

Calculate Capacitance of Each Section

The capacitance \( C \) of a parallel-plate capacitor with a dielectric constant \( K \) is given by:\[ C = \frac{\epsilon_0 K A}{d} \]For each section, since both have thickness \( \frac{d}{2} \), the capacitances are:\[ C_1 = \frac{2\epsilon_0 K_1 A}{d} \]\[ C_2 = \frac{2\epsilon_0 K_2 A}{d} \]
04

Combine Capacitance in Series

Applying the formula for capacitors in series:\[ \frac{1}{C} = \frac{d}{2\epsilon_0 K_1 A} + \frac{d}{2\epsilon_0 K_2 A} \]Simplifying gives:\[ \frac{1}{C} = \frac{d(K_1 + K_2)}{2\epsilon_0 K_1 K_2 A} \]Therefore, the capacitance is:\[ C = \frac{2\epsilon_0 A}{d} \left( \frac{K_1 K_2}{K_1 + K_2} \right) \]
05

Check for Special Cases

Case (i): If only one dielectric with constant \( K \) is present, the formula becomes:\[ C = \frac{\epsilon_0 K A}{d} \]Case (ii): If filled with air (vacuum), \( K = 1 \):\[ C = \frac{\epsilon_0 A}{d} \]Both cases agree with known results, verifying our formula.

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Key Concepts

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

Dielectrics
Dielectrics are insulating materials that improve the efficiency of capacitors by increasing their ability to store charge. When inserted into a capacitor, a dielectric increases its capacitance by reducing the electric field inside the capacitor for a given charge on the plates.

### How Dielectrics WorkDielectrics are characterized by a property known as the dielectric constant (\( K \)). The dielectric constant is a measure of a material's ability to concentrate electric flux, and each material has its own specific value.
- When placed between the plates of a capacitor, the dielectric is polarized.- The polarization creates an induced electric field that opposes the field created by the charge on the plates, effectively reducing the net electric field in the capacitor.
- This means for the same charge, a dielectric allows for a smaller voltage across the plates, thereby increasing the capacitance.
In our exercise, the use of two different dielectrics, each with thickness \(d/2\), highlights how multiple materials can be combined to manipulate capacitance. The result is a combined effect that must consider the properties of each material.
Capacitance
Capacitance is a measure of a capacitor's capacity to store an electric charge. It is defined as the ratio of the electric charge on each conductor to the potential difference between them.

### Calculating CapacitanceThe capacitance \( C \) of a parallel-plate capacitor filled with a dielectric is given by:\[ C = \frac{\epsilon_0 K A}{d} \]Where:
  • \( \epsilon_0 \) is the permittivity of free space.
  • \( K \) is the dielectric constant of the material.
  • \( A \) is the area of one of the plates.
  • \( d \) is the distance between the two plates.
In a mixed dielectric material scenario, such as the exercise involving two dielectrics \(K_1\) and \(K_2\), the capacitance is not simply the sum of capacities of individual sections. Instead, the behavior of capacitors in parallel and series must be considered, as we'll explore next.
Capacitors in Series
When capacitors are connected in series, their total capacitance is not the simple sum of their individual capacitances. Instead, the reciprocal of the total capacitance \(C_{total}\) is equal to the sum of the reciprocals of the individual capacitances. This concept is critical in solving the exercise, where the capacitor is considered to have two sections in series.

### Formula for Series CapacitorsFor two capacitors \( C_1 \) and \( C_2 \) connected in series, the total capacitance \( C_{total} \) is calculated using:\[ \frac{1}{C_{total}} = \frac{1}{C_1} + \frac{1}{C_2} \]In our exercise, each section of the capacitor is treated like a separate capacitor filled with its own dielectric, following:- Section 1 with capacitance \(C_1 = \frac{2\epsilon_0 K_1 A}{d}\)- Section 2 with capacitance \(C_2 = \frac{2\epsilon_0 K_2 A}{d}\)Applying the series formula gives the overall capacitance, ensuring that the effects of both dielectrics are accounted for properly.

Understanding this principle helps clarify how a mixed dielectric system behaves, and illustrates the nuanced approach needed when combining different materials within one capacitor.

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Most popular questions from this chapter

How far from a \(-7.20 \mu \mathrm{C}\) point charge must a \(+2.30 \mu \mathrm{C}\) point charge be placed in order for the electric potential energy of the pair of charges to be \(-0.400 \mathrm{J} ?\) (Take the energy to be zero when the charges are infinitely far apart.)

Electronic flash units for cameras contain a capacitor for storing the energy used to produce the flash. In one such unit, the flash lasts for \(\frac{1}{675}\) s with an average light power output of \(2.70 \times 10^{5} \mathrm{W}\) (a) If the conversion of electrical energy to light is 95\(\%\) efficient (the rest of the energy goes to thermal energy), how much energy must be stored in the capacitor for one flash? (b) The capacitor has a potential difference between its plates of 125 \(\mathrm{V}\) when the stored energy equals the value calculated in part (a). What is the capacitance?

A parallel-plate air capacitor is made from two plates 0.200 m square, spaced 0.800 \(\mathrm{cm}\) apart. It is connected to a \(120-\mathrm{V}\) battery. (a) What is the capacitance? (b) What is the charge on each plate? (c) What is the electric field between the plates? (d) What is the energy stored in the capacitor? (e) If the battery is disconnected and then the plates are pulled apart to a separation of \(1.60 \mathrm{cm},\) what are the answers to parts (a), (b), \((c),\) and \((d) ?\)

A parallel-plate air capacitor has a capacitance of 920 pF. The charge on each plate is 2.55\(\mu \mathrm{C}\) . (a) What is the potential difference between the plates? (b) If the charge is kept constant, what will be the potential difference between the plates if the separation is doubled? (c) How much work is required to double the separation?

A parallel-plate capacitor has capacitance \(C_{0}=5.00 \mathrm{pF}\) when there is air between the plates. The separation between the plates is 1.50 \(\mathrm{mm}\) (a) What is the maximum magnitude of charge \(Q\) that can be placed on each plate if the electric field in the region between the plates is not to exceed \(3.00 \times 10^{4} \mathrm{V} / \mathrm{m}\) (b) A dielectric with \(K=2.70\) is inserted between the plates of the capacitor, completely filling the volume between the plates. Now what is the maximum magnitude of charge on each plate if the electric field between the plates is not to exceed \(3.00 \times 10^{4} \mathrm{V} / \mathrm{m} ?\)

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