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A parallel-plate capacitor is to be constructed by using, as a dielectric, rubber with a dielectric constant of 3.20 and a dielectric strength of 20.0 \(\mathrm{MV} / \mathrm{m}\) . The capacitor is to have a capacitance of 1.50 \(\mathrm{nF}\) and must be able to withstand a maximum potential difference of 4.00 \(\mathrm{kV} .\) What is the minimum area the plates of this capacitor can have?

Short Answer

Expert verified
The minimum area the plates can have is approximately 0.106 m².

Step by step solution

01

Identify Key Formulas

We need to use two formulas to find the minimum area of the plates: the formula for the capacitance of a parallel-plate capacitor \( C = \frac{\varepsilon_0 \varepsilon_r A}{d} \), where \( \varepsilon_0 \) is the vacuum permittivity \( 8.85 \times 10^{-12} \ \text{F/m} \), \( \varepsilon_r \) is the dielectric constant (3.20), \( A \) is the area, and \( d \) is the separation distance. The second formula is the dielectric strength \( V = E \cdot d \), where \( E \) is the maximum electric field and \( V \) is the potential difference.
02

Calculate Separation Distance

Rearrange the dielectric strength formula to find the separation distance \( d \). The dielectric strength given is \( 20.0 \ \text{MV/m} \) or \( 20.0 \times 10^6 \ \text{V/m} \). The maximum potential difference \( V \) is \( 4.00 \ \text{kV} \) or \( 4.00 \times 10^3 \ \text{V} \). So we have: \[ d = \frac{V}{E} = \frac{4.00 \times 10^3}{20.0 \times 10^6} = 2.00 \times 10^{-4} \ \text{m} \].
03

Solve for Required Area

Substituting \( C = 1.50 \times 10^{-9} \ \text{F} \), \( \varepsilon_r = 3.20 \), and \( d = 2.00 \times 10^{-4} \ \text{m} \) into the capacitance formula, we get: \[ 1.50 \times 10^{-9} = \frac{8.85 \times 10^{-12} \times 3.20 \times A}{2.00 \times 10^{-4}} \]Solve for \( A \):\[ A = \frac{(1.50 \times 10^{-9}) \times (2.00 \times 10^{-4})}{8.85 \times 10^{-12} \times 3.20} \approx 0.106 \ \text{m}^2 \]
04

Verify Plate Area

For full verification, ensure \( A \) calculated is large enough for the capacitor to handle the given potential difference and capacitance requirements without excessive electric field providing another check: \( d = 2.00 \times 10^{-4} \ \text{m} \) and \( A \approx 0.106 \ \text{m}^2 \) are consistent with capacitance and dielectric strength under the constraints.

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Key Concepts

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

Dielectric Constant
The dielectric constant, also known as the relative permittivity, is a number that describes how much an electric field is reduced inside a dielectric material compared to a vacuum. In the case of capacitors, a higher dielectric constant means that the material can store more electrical energy for a given electric field.

For our rubber dielectric in the original problem, the dielectric constant is 3.20. This means that rubber can store 3.20 times more electrical energy than a vacuum for the same electric field. Also, using materials with higher dielectric constants allows for more compact capacitors with more capacitance.

When designing capacitors, choosing materials with suitable dielectric constants is essential to meet size and performance requirements. This parameter predominantly affects the capacitance, which influences how much charge the plates can store.
Dielectric Strength
Dielectric strength measures the maximum electric field a material can withstand without experiencing breakdown, which is when it starts to conduct electricity. For rubber, the dielectric strength in the original problem is given as 20.0 MV/m. This tells us that the material can sustain an electric field as strong as 20 million volts per meter before failing.

The dielectric strength is critical when determining the separation distance between the plates in a capacitor. It dictates how closely the plates can be positioned without causing the dielectric to break down. A higher dielectric strength means a larger maximum voltage can be applied across the capacitor without causing failure, making the material very valuable in high-voltage applications.
Parallel-Plate Capacitor
A parallel-plate capacitor is a simple design featuring two conductive plates separated by a dielectric material. This setup is popular due to its simplicity, allowing uniform electric fields and predictable capacitance. In the context of the original exercise, a capacitor is constructed with rubber as the dielectric between two plates.

The distance between the plates and the area of the plates are crucial factors; they determine the capacitor's ability to store charge. The capacitance is directly proportional to the plate area and the dielectric constant, and inversely proportional to the distance between the plates. This design offers convenience in manipulation by adjusting these variables to achieve the desired capacitance to meet specific application requirements.
Electric Field
The electric field in a capacitor is the force experienced by a charge situated within the electric field created between the capacitor's plates. In the context of our problem, it is represented in the relationship for dielectric strength: \[ V = E \cdot d \]Here, \( V \) is the voltage difference, \( E \) is the electric field strength, and \( d \) is the distance between the plates.

Within a parallel-plate capacitor using a uniform dielectric like rubber, the electric field is uniform between the plates. This uniformity helps in maintaining consistent capacitor performance. The formula shows how a smaller gap \( d \) increases the electric field for a given voltage, stressing the importance of using materials with high dielectric strength to prevent breakdown.
Capacitance Formula
The capacitance formula is fundamental to understanding how to design and evaluate capacitors: \[ C = \frac{\varepsilon_0 \varepsilon_r A}{d} \]In this formula, \( C \) is the capacitance, \( \varepsilon_0 \) is the permittivity of free space, \( \varepsilon_r \) the dielectric constant, \( A \) the plate area, and \( d \) the separation distance between plates.

For the problem at hand, the given capacitance is 1.50 nF. By rearranging the formula, designers can solve for the minimum required plate area, ensuring that the dielectric strength and specified voltage ratings are not exceeded. This formula highlights the direct relationship between capacitance, plate size, and dielectric properties, providing a clear path for engineers to tailor capacitors for their intended use without exceeding material limits.

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Most popular questions from this chapter

At a certain distance from a point charge, the potential and electric-field magnitude due to that charge are 4.98 V and \(12.0 \mathrm{V} / \mathrm{m},\) respectively. (Take the potential to be zero at infinity.) (a) What is the distance to the point charge? (b) What is the magnitude of the charge? (c) Is the electric field directed toward or away from the point charge?

X-ray tube. An X-ray tube is similar to a cathode-ray tube. (See previous problem.) Electrons are accelerated to high speeds at one end of the tube. If they are moving fast enough when they hit the target at the other end, they give up their energy as X-rays (a form of nonvisible light). (a) Through what potential difference should electrons be accelerated so that their speed is 1.0\(\%\) of the speed of light when they hit the target? (b) What potential difference would be needed to give protons the same kinetic energy as the electrons? (c) What speed would this potential difference give to protons? Express your answer in \(\mathrm{m} / \mathrm{s}\) and as a percent of the speed of light.

A small particle has charge \(-5.00 \mu \mathrm{C}\) and mass \(2.00 \times\) \(10^{-4} \mathrm{kg} .\) It moves from point \(A,\) where the electric potential is \(V_{A}=+200 \mathrm{V},\) to point \(B,\) where the electric potential is \(V_{B}=+800 \mathrm{V} .\) The electric force is the only force acting on the particle. The particle has speed 5.00 \(\mathrm{m} / \mathrm{s}\) at point \(A .\) What is its speed at point \(B ?\) Is it moving faster or slower at \(B\) than at \(A\)? Explain.

A parallel-plate capacitor is made from two plates 12.0 \(\mathrm{cm}\) on each side and 4.50 mm apart. Half of the space between these plates contains only air, but the other half is filled with Plexiglas\oplus of dielectric constant 3.40 . (See Figure \(18.55 .\) An 18.0 V battery is connected across the plates. (a) What is the capacitance of this combination? (Hint: Can you think of this capacitor as equivalent to two capacitors in parallel? (b) How much energy is stored in the capacitor? (c) If we remove the Plexiglas@, but change nothing else, how much energy will be stored in the capacitor?

A particle with a charge of \(+4.20 \mathrm{nC}\) is in a uniform electric field \(\vec{\boldsymbol{E}}\) directed to the left. It is released from rest and moves to the left; after it has moved \(6.00 \mathrm{cm},\) its kinetic energy is found to be \(+1.50 \times 10^{-6} \mathrm{J}\) (a) What work was done by the electric force? (b) What is the potential of the starting point with respect to the endpoint? (c) What is the magnitude of \(\vec{E} ?\)

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