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Chapter 24: Problem 3

24.3. A parallel-plate air capacitor of capacitance 245 \(\mathrm{pF}\) has a charge of magnitude 0.148\(\mu \mathrm{C}\) on each plate. The plates are 0.328 \(\mathrm{mm}\) apart. (a) What is the potential difference between the plates? (b) What is the area of each plate? (c) What is the electric-field magnitude between the plates? (d) What is the surface charge density on each plate?

Short Answer

Expert verified
(a) 603.27 V; (b) 9.07 \times 10^{-3} m^2; (c) 1.84 \times 10^{6} V/m; (d) 1.63 \times 10^{-5} C/m^2.

Step by step solution

01

Find the Potential Difference

The potential difference \( V \) between the plates of a capacitor can be found using the formula for capacitance: \( C = \frac{Q}{V} \), where \( C \) is the capacitance, \( Q \) is the charge, and \( V \) is the potential difference. Rearranging gives \( V = \frac{Q}{C} \). Plug in \( Q = 0.148 \, \mu \mathrm{C} = 0.148 \times 10^{-6} \, \mathrm{C} \) and \( C = 245 \, \mathrm{pF} = 245 \times 10^{-12} \, \mathrm{F} \):\[ V = \frac{0.148 \times 10^{-6}}{245 \times 10^{-12}} = 603.27 \, \mathrm{V} \]
02

Calculate the Area of Each Plate

The area \( A \) of each plate can be found using the formula for capacitance in a parallel-plate capacitor: \( C = \varepsilon_0 \frac{A}{d} \), where \( \varepsilon_0 = 8.85 \times 10^{-12} \, \mathrm{F/m} \) is the permittivity of free space and \( d = 0.328 \, \mathrm{mm} = 0.328 \times 10^{-3} \, \mathrm{m} \) is the separation between the plates. Rearranging for \( A \):\[ A = C \frac{d}{\varepsilon_0} \]Substitute the known values:\[ A = 245 \times 10^{-12} \times \frac{0.328 \times 10^{-3}}{8.85 \times 10^{-12}} = 9.07 \times 10^{-3} \, \mathrm{m}^2 \]
03

Calculate the Electric Field Magnitude

The electric field magnitude \( E \) between the plates can be calculated by \( E = \frac{V}{d} \). We already found \( V = 603.27 \, \mathrm{V} \) and \( d = 0.328 \times 10^{-3} \, \mathrm{m} \).\[ E = \frac{603.27}{0.328 \times 10^{-3}} = 1.84 \times 10^{6} \, \mathrm{V/m} \]
04

Calculate the Surface Charge Density

The surface charge density \( \sigma \) can be calculated as \( \sigma = \frac{Q}{A} \). We know \( Q = 0.148 \times 10^{-6} \, \mathrm{C} \) and from Step 2, \( A = 9.07 \times 10^{-3} \, \mathrm{m}^2 \).\[ \sigma = \frac{0.148 \times 10^{-6}}{9.07 \times 10^{-3}} = 1.63 \times 10^{-5} \, \mathrm{C/m}^2 \]

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Key Concepts

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

Electric Field
An electric field represents a force per unit charge exerted around charged objects. When dealing with capacitor plates, the electric field is key in understanding how forces act on charges between the plates. The electric field strength, denoted as \( E \), between parallel capacitor plates can be found using the formula:
  • \( E = \frac{V}{d} \)
where \( V \) is the potential difference across the plates and \( d \) is the distance between them. For parallel plates, the field is uniform, meaning it has the same strength and direction at every point between the plates.
In this context, knowing that the potential difference is 603.27 V and the distance between the plates is 0.328 mm, we calculated the electric field magnitude to be \( E = 1.84 \times 10^6 \) V/m. This high field indicates strong force interactions along the gap between plates, crucial for the capacitor's ability to store charges.
Potential Difference
The potential difference, often referred to as voltage, between two points is the work done to move a charge from one point to another. In capacitors, this potential difference pushes and holds electric charges on the plates.
To find the potential difference \( V \) in a capacitor, the formula used is:
  • \( V = \frac{Q}{C} \)
where \( Q \) is the charge on the plates and \( C \) is the capacitance of the capacitor. In our exercise, by substituting the known values \( Q = 0.148 \times 10^{-6} \) C and \( C = 245 \times 10^{-12} \) F into this formula, we calculated \( V \) to be approximately 603.27 V. This voltage indicates the energy required to move a charge across the plates, and is an essential quantity in assessing how capacitors will perform in circuits.
Surface Charge Density
Surface charge density, symbolized as \( \sigma \), describes how much charge resides on a unit area of a surface. It's a crucial property for understanding charge distribution on plates, affecting their electric field and potential.
For capacitors, surface charge density can be determined by the equation:
  • \( \sigma = \frac{Q}{A} \)
where \( Q \) is the charge and \( A \) is the area of the plate. In our problem, with a charge \( Q = 0.148 \times 10^{-6} \) C and an area \( A = 9.07 \times 10^{-3} \) m虏, the surface charge density was computed to be \( 1.63 \times 10^{-5} \) C/m虏. This value indicates how much charge is spread over each square meter of the plate, directly impacting the electric field's intensity between the plates. Understanding this helps in designing capacitors with desired properties for specific electronic applications.

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Most popular questions from this chapter

24.25. A \(5.80-\mu F\) , parallel-plate, air capacitor has a plate separation of 5.00 \(\mathrm{mm}\) and is charged to a potential difference of 400 \(\mathrm{V}\) . Calculate the energy density in the region between the plates, in units of \(\mathrm{J} / \mathrm{m}^{3}\) .

24.8. A \(5.00-\) pF, parallel-plate, air-filled capacitor with circular plates is to be used in a circuit in which it will be subjected to potentials of up to \(1.00 \times 10^{2} \mathrm{V}\) . The electric field between the plates is to be no greater than \(1.00 \times 10^{4} \mathrm{N} / \mathrm{C}\) . As a budding electrical engineer for Live-Wire Electronics, your tasks are to (a) design the capacitor by finding what its physical dimensions and separation must be; (b) find the maximum charge these plates can hold.

24.74. The parallel-plate air capacitor in Fig. 24.40 consists of two horizontal conducting plates of equal area \(A .\) The bottom plate rests on a fixed support, and the top plate is suspended by four springs with spring constant \(k,\) positioned at each of the four corners of the top plate as shown in the figure. When uncharged, the plates are separated by a distance \(z_{0}\) . A battery is connected to the plates and produces a potential difference \(V\) between them. This causes the plate separation to decrease to \(z\) . Neglect any fringing effects. (a) Show that the electrostatic force between the charged plates has a magnitude \(\epsilon_{0} A V^{2} / 2 z^{2}\) . (Hint: See Exercise \(24.29 . )\) (b) Obtain an expression that relates the plate separation \(z\) to the potential difference \(V\) . The resulting equation will be cubic in \(z\) . (c) Given the values \(A=0.300 \mathrm{m}^{2}, z_{0}=1.20 \mathrm{mm}, k=25.0 \mathrm{N} / \mathrm{m},\) and \(V=120 \mathrm{V},\) find the two values of \(z\) for which the top plate will be in equilibrium. (Hint: You can solve the cubic equation by plugging a trial value of \(z\) into the equation and then adjusting your guess until the equation is satisfied to three significant figures. Locating the roots of the cubic equation graphically can help you pick starting values of \(z\) for this trial-and error procedure. One root of the cubic equation has a nonphysical negative value. (d) For each of the two values of \(z\) found in part (c), is the equilibrium stable or unstable? For stable equilibrium a small displacement of the object will give rise to a net force tending to return the object to the equilibrium position. For unstable equilibrium a small displacement gives rise to a net force that takes the object farther away from equilibrium.

24.58. Several \(0.25-\mu \mathrm{F}\) capacitors are available. The voltage across each is not to exceed 600 \(\mathrm{V}\) . You need to make a capacitor with capacitance 0.25\(\mu \mathrm{F}\) to be connected across a potential difference of 960 \(\mathrm{V}\) . (a) Show in a diagram how an cquivalent capacitor with the desired properties can be obtained. (b) No dielectric is a perfect insulator that would not permit the flow of any charge through its volume. Suppose that the dielectric in one of the capacitors in your diagram is a moderately good conductor. What will happen in this case when your combination of capacitors is connected across the \(960-\mathbf{V}\) potential difference?

24.47. A \(12.5-\mu F\) capacitor is connected to a power supply that keeps a constant potential difference of 24.0 \(\mathrm{V}\) across the plates. A piece of material having a dielectric constant of 3.75 is placed between the plates, completely filling the space between them. (a) How much energy is stored in the capacitor before and after the diclectric is inserted? (b) By how much did the energy change dur- ing the insertion? Did it increase or decrease?
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