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(II) An electric field of \(4.80 \times 10^{5} \mathrm{V} / \mathrm{m}\) is desired between two parallel plates, each of area 21.0 \(\mathrm{cm}^{2}\) and separated by 0.250 \(\mathrm{cm}\) of air. What charge must be on each plate?

Short Answer

Expert verified
Each plate must have a charge of approximately \(8.93 \times 10^{-9}\) C.

Step by step solution

01

Understand the Problem and the Given Values

The problem involves two parallel plates where an electric field of \(4.80 \times 10^{5} \mathrm{V}/\mathrm{m}\) is desired. The plates have an area of \(21.0 \ \mathrm{cm}^{2}\) and are separated by a distance of \(0.250 \ \mathrm{cm}\). We need to find the charge that must be placed on each plate to achieve this electric field.
02

Convert Units

Convert the given units into SI units. The area is given in \(\mathrm{cm}^{2}\), which needs to be converted to \(\mathrm{m}^{2}\): \[ A = 21.0 \ \mathrm{cm}^{2} = 21.0 \times 10^{-4} \ \mathrm{m}^{2} \] Similarly, convert the separation distance from \(\mathrm{cm}\) to \(\mathrm{m}\): \[ d = 0.250 \ \mathrm{cm} = 0.00250 \ \mathrm{m} \]
03

Use the Relationship Between Electric Field and Charge

The electric field \(E\) between two parallel plates is related to the voltage \(V\) and the plate separation \(d\) by the equation \(E = \frac{V}{d}\), and can also be related to surface charge density \(\sigma\) by \(E = \frac{\sigma}{\varepsilon_0}\), where \(\varepsilon_0\) is the permittivity of free space, \(\varepsilon_0 = 8.85 \times 10^{-12} \ \mathrm{C}^{2}/\mathrm{N}\mathrm{m}^{2}\). Simply rearrange this to find \(\sigma\): \[ \sigma = E \cdot \varepsilon_0 \]
04

Calculate the Surface Charge Density

Substitute the known values into the equation: \[ \sigma = (4.80 \times 10^{5} \ \mathrm{V}/\mathrm{m}) \times (8.85 \times 10^{-12} \ \mathrm{C}^{2}/\mathrm{N}\mathrm{m}^{2}) \] Calculate \(\sigma\).
05

Find the Total Charge on Each Plate

Once \(\sigma\) is found, use the area \(A\) to calculate the total charge \(Q\) on each plate:\[ Q = \sigma \times A \] Substitute \(\sigma\) from the previous step and \(A = 21.0 \times 10^{-4} \ \mathrm{m}^{2}\) into the equation and calculate \(Q\).
06

Calculate the Final Charge Value

Use the values obtained to compute \(Q\): \[ Q = (\sigma) \times (21.0 \times 10^{-4} \ \mathrm{m}^{2}) \]Find the numerical value for \(Q\), which represents the charge needed on each plate.

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Key Concepts

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

Parallel Plates
Parallel plates are a fundamental concept in understanding electric fields. They are essentially two large, flat sheets placed parallel to each other. When a voltage is applied across these plates, an electric field is created between them. This field is uniform, meaning its strength is constant throughout the space between the plates. This makes parallel plates an ideal setup for experiments and calculations in electrostatics.

In physics, this uniform electric field is determined by the potential difference between the plates and the distance separating them. To conceptualize, imagine the plates as two large charged surfaces facing each other with a small gap in between. If one plate holds positive charge and the other negative, the electric field lines are straight and parallel, with a consistent direction from the positive to the negative plate.

This setup is widely used in various applications, such as capacitors, which store electrical energy. Understanding the behavior of electric fields between parallel plates is crucial for engineers and physicists when designing circuits and electronic components.
Surface Charge Density
Surface charge density, denoted by the Greek letter \(\sigma\), is a measure of how much electric charge resides on a surface. It is defined as the amount of charge per unit area of the plates. This is particularly important in the context of parallel plates, as it directly influences the electric field generated between the plates.

To calculate surface charge density in the scenario of parallel plates, use the formula \(\sigma = \frac{Q}{A}\), where \(Q\) represents the total charge on the plate and \(A\) is the area of the plate. This equation reveals how charges spread uniformly across the plate surface.

Moreover, surface charge density relates to the electric field by the equation \(E = \frac{\sigma}{\varepsilon_0}\), which ties the strength of the electric field to the charge density and the permittivity of free space, \(\varepsilon_0\). Thus, as \(\sigma\) increases for a given permittivity, so does the electric field strength. Understanding surface charge density is key to manipulating electric fields in practical applications.
Permittivity of Free Space
The permittivity of free space, represented as \(\varepsilon_0\), is a fundamental constant in electromagnetism. It quantifies the ability of a vacuum to permit electric field lines. Essentially, \(\varepsilon_0\) describes how much resistance is encountered when forming an electric field in free space.

This constant plays a vital role when calculating electric fields, especially between parallel plates. The relationship \(E = \frac{\sigma}{\varepsilon_0}\) highlights how \(\varepsilon_0\) impacts the field strength. Here, \(\varepsilon_0\) is approximately \(8.85 \times 10^{-12} \mathrm{C}^{2}/\mathrm{N}\mathrm{m}^{2}\). It ensures that electric field equations are balanced by taking into account the medium's ability to conduct electric flux.

Understanding \(\varepsilon_0\) is crucial because it is a baseline against which other materials' permittivities are compared. This constant provides insight into how different materials affect electric fields. Recognizing its application helps in designing reliable electronic devices and circuits by ensuring the electric fields are managed effectively.

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Most popular questions from this chapter

Determine the capacitance of the Earth, assuming it to be a spherical conductor.

A \(2.70-\mu \mathrm{F}\) capacitor is charged to \(475 \mathrm{~V}\) and a \(4.00-\mu \mathrm{F}\) capacitor is charged to \(525 \mathrm{~V}\). \((a)\) These capacitors are then disconnected from their batteries, and the positive plates are now connected to each other and the negative plates are connected to each other. What will be the potential difference across each capacitor and the charge on each? (b) What is the voltage and charge for each capacitor if plates of opposite sign are connected?

Two capacitors connected in parallel produce an equivalent capacitance of \(35.0 \mu \mathrm{F}\) but when connected in series the equivalent capacitance is only \(5.5 \mu \mathrm{F}\). What is the individual capacitance of each capacitor?

A parallel-plate capacitor with plate area \(2.0 \mathrm{~cm}^{2}\) and airgap separation \(0.50 \mathrm{~mm}\) is connected to a \(12-\mathrm{V}\) battery, and fully charged. The battery is then disconnected. ( \(a\) ) What is the charge on the capacitor? (b) The plates are now pulled to a separation of \(0.75 \mathrm{~mm}\). What is the charge on the capacitor now? (c) What is the potential difference across the plates now? ( \(d\) ) How much work was required to pull the plates to their new separation?

Consider three capacitors, of capacitance \(3600 \mathrm{pF}\) \(5800 \mathrm{pF},\) and \(0.0100 \mu \mathrm{F}\). What maximum and minimum capacitance can you form from these? How do you make the connection in each case?

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