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Chapter 18: Problem 28

A parallel-plate capacitor having plates 6.0 \(\mathrm{cm}\) apart is connected across the terminals of a 12 \(\mathrm{V}\) battery. (a) Being as quantitative as you can, describe the location and shape of the equipotential surface that is at a potential of \(+6.0 \mathrm{V}\) relative to the potential of the negative plate. Avoid the edges of the plates. (b) Do the same for the equipotential surface that is at \(+2.0 \mathrm{V}\) relative to the negative plate. (c) What is the potential gradient between the plates?

Short Answer

Expert verified
Equipotential surfaces: +6.0 V at 3.0 cm, +2.0 V at 1.0 cm from negative plate. Potential gradient: 200 V/m.

Step by step solution

01

Understand the Configuration

The capacitor consists of two parallel plates separated by a distance of 6.0 cm. The negative plate is connected to the ground (0 V), and the positive plate is at +12.0 V.
02

Calculate the Electric Field

The electric field between two parallel plates is uniform and given by the potential difference divided by the separation distance.\[ E = \frac{V}{d} = \frac{12 \, \text{V}}{0.06 \, \text{m}} = 200 \, \text{V/m} \]
03

Describe the +6.0 V Equipotential Surface

The potential of +6.0 V is halfway between 0 V and +12.0 V, and this equipotential surface is a plane parallel to the plates, located halfway between the plates. Since the plates are 6.0 cm apart, this surface is 3.0 cm from each plate.
04

Describe the +2.0 V Equipotential Surface

The +2.0 V equipotential surface is closer to the negative plate. Using the potential gradient, find the distance:\[ V = Ed \Rightarrow d = \frac{V}{E} = \frac{2.0 \, \text{V}}{200 \, \text{V/m}} = 0.01 \, \text{m} \] The equipotential surface is 1.0 cm from the negative plate.
05

Calculate the Potential Gradient

The potential gradient is equivalent to the electric field strength, which is the rate of change of potential with respect to distance:\[ \text{Potential Gradient} = V/d = 200 \, \text{V/m} \]

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Key Concepts

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

Equipotential Surfaces
In a parallel-plate capacitor, equipotential surfaces are imaginary surfaces where the electrical potential is constant throughout. These surfaces are crucial for understanding how electric fields interact within a capacitor. In this setup:
  • Equipotential surfaces are parallel to the plates of the capacitor.
  • There is no potential difference across any point on an equipotential surface.
  • The +6 V surface is located halfway between the plates, which are 6 cm apart, meaning it lies precisely 3 cm from either plate.
  • The +2 V surface is closer to the negative plate, exactly 1 cm away from it.
Equipotential surfaces help visualize potential energy distribution, showing how charge resides and how work is done in moving charge across different potentials.
Potential Gradient
The potential gradient refers to the change in electrical potential between two points divided by the distance separating those points. This is a critical concept in understanding capacitors and electric fields.
  • In a parallel-plate capacitor, the potential gradient is constant because the electric field is uniform.
  • The potential gradient equals the electric field strength. It is calculated as the potential difference divided by the separator's distance.
  • In our example, with a potential difference of 12 V and a plate separation of 6 cm (or 0.06 m), the potential gradient is 200 V/m.
A uniform potential gradient means that the electric field exerts the same force on a charge throughout the space between the plates. This uniformity allows us to predict behavior within the capacitor with greater accuracy and certainty.
Electric Field
The electric field in a parallel-plate capacitor is a vital element in understanding how capacitors function. Between the plates, the electric field:
  • Is uniform and directed from the positive to the negative plate.
  • Can be calculated using the formula \( E = \frac{V}{d} \), with \( V \) as the potential difference and \( d \) as the separation distance.
  • Equals 200 V/m for this setup, showing how strong the field is between the plates.
A uniform electric field means that a test charge placed anywhere between the plates experiences the same force throughout. This uniformity influences numerous applications and is key in designing various electrical components.
Parallel-Plate Capacitor
The parallel-plate capacitor is a basic electronic component used for storing electrical energy. It consists of two conductive plates separated by a dielectric (non-conductive) material.
  • It functions based on the principle of storing energy in an electric field created between the plates.
  • The capacitance, \( C \), of a parallel-plate capacitor is determined by the formula \( C = \varepsilon \frac{A}{d} \), with \( \varepsilon \) representing the permittivity of the dielectric, \( A \) the area of the plates, and \( d \) the distance between them.
  • The larger the area of the plates or the smaller the distance between them, the greater the capacitance.
Understanding the dynamics of parallel-plate capacitors helps in designing circuits and managing electronic systems, making them indispensable in electronics.

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Most popular questions from this chapter

Electric eels. Electric eels and electric fish generate large potential differences that are used to stun 9 enemies and prey. These potentials are produced by cells that each can generate 0.10 V. We can plausibly model such cells as charged capacitors. (a) How should these cells be connected \((\mathrm{in}\) series or in parallel) to produce a total potential of more than 0.10 \(\mathrm{V} ?\) (b) Using the connection in part (a), how many cells must be connected together to produce the 500 \(\mathrm{V}\) surge of the electric eel?

(a) You find that if you place charges of \(\pm 1.25 \mu \mathrm{C}\) on two separated metal objects, the potential difference between them is 11.3 \(\mathrm{V}\) . What is their capacitance? (b) A capacitor has a capacitance of 7.28\(\mu \mathrm{F}\) . What amount of excess charge must be placed on each of its plates to make the potential difference between the plates equal to 25.0 \(\mathrm{V}\) ?

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.

A metal sphere carrying an evenly distributed charge will have spherical equipotential surfaces surrounding it. Suppose the sphere's radius is 50.0 \(\mathrm{cm}\) and it carries a total charge of \(+1.50 \mu \mathrm{C}\) (a) Calculate the potential of the sphere's surface.(b) You want to draw equipotential surfaces at intervals of 500 \(\mathrm{V}\) outside the sphere's surface. Calculate the distance between the first and the second equipotential surfaces, and between the 20 \(\mathrm{th}\) and 21 \(\mathrm{st}\) equipotential surfaces. (c) What does the changing spacing of the surfaces tell you about the electric field?

You are working on an electronics project requiring a variety of capacitors, but have only a large supply of 100 nF capacitors available. Show how you can connect these capacitors to produce each of the following equivalent capacitances: (a) \(50 \mathrm{nF},\) (b) \(450 \mathrm{nF},(\mathrm{c}) 25 \mathrm{nF},\) (d) 75 \(\mathrm{nF.}\)
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