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Chapter 23: Problem 45

A potential difference of 480 \(\mathrm{V}\) is established between large, parallel, metal plates. Let the potential of one plate be 480 \(\mathrm{V}\) and the other be 0 \(\mathrm{V}\) . The plates are separated by \(d=1.70 \mathrm{cm} .\) (a) Sketch the equipotential surfaces that correspond to \(0,120,\) \(240,360,\) and 480 \(\mathrm{V}\) . (b) In your sketch, show the electric field lines. Does your sketch confirm that the field lines and equipotential surfaces are mutually perpendicular?

Short Answer

Expert verified
Equipotential surfaces are evenly spaced at 0.425 cm apart; electric field lines are perpendicular to these surfaces.

Step by step solution

01

Understand Equipotential Surfaces

Equipotential surfaces are surfaces on which every point has the same electric potential. Within a uniform electric field, equipotential surfaces between parallel plate capacitors are planes parallel to the plates. For a potential difference of 480 V between the plates, the equipotential surfaces will be evenly distributed because the field is uniform.
02

Find the Distance Between Equipotential Surfaces

Given that the potential difference across the plates is 480 V and the distance between the plates, \(d\), is 1.70 cm, we divide the total potential by the number of equal divisions (5 in this case: 0, 120, 240, 360, and 480 V) to find the potential separation in volts of each surface (120 V). Calculate the distance between each equipotential surface as: \(\Delta d = \frac{d}{4}\). Hence, \[ \Delta d = \frac{1.70 \text{ cm}}{4} = 0.425 \text{ cm}. \]
03

Draw the Equipotential Surfaces

Using \(\Delta d = 0.425 \text{ cm}\), draw lines parallel to the plates at equal distances from one another. Starting from the 0 V plate, draw lines at 0.425 cm intervals until reaching the 480 V plate to mark the surfaces of 120 V, 240 V, and 360 V.
04

Draw Electric Field Lines

Electric field lines in a parallel plate capacitor are straight and perpendicular to the plates. They go from the positive plate (480 V) directly to the negative plate (0 V). Draw these lines perpendicular to the equipotential surfaces which confirms that field lines are perpendicular to the equipotential surfaces.
05

Confirm Perpendicularity

Ensure each electric field line crosses the equipotential surfaces at a right angle. If all lines are perpendicular, this demonstrates that electric field lines and equipotential surfaces are mutually perpendicular, confirming the principles of electric fields.

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Key Concepts

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

Equipotential Surfaces
Equipotential surfaces are fascinating concepts in electrostatics. They are imaginary surfaces over which the electric potential is the same at every point. Imagine walking across a perfectly even floor, where every step keeps you at the same altitude; this is similar to moving along an equipotential surface.
These surfaces are instrumental in understanding electric fields because they indicate areas where no work is performed when moving a charge. Since the potential doesn't change on these surfaces, the work done is zero.
  • In a highly uniform field, like between the plates of a parallel plate capacitor, equipotential surfaces are evenly spaced planes parallel to the plates.
  • In our example, with a potential difference of 480 V, these surfaces are distributed at intervals calculated by dividing the total potential difference into fractions, depending on how many equipotential markings you need (5 in this scenario).
They help visualize the electric potential's spatial variation without the need for detailed calculations.
Parallel Plate Capacitor
A parallel plate capacitor is a simple yet powerful device used to store electric charge. It consists of two metal plates placed parallel to each other, separated by some distance.
The physics behind this setup is straightforward but profound, as it exemplifies two core concepts: uniform electric fields and potential difference.
  • The uniform electric field between the plates can be calculated using the potential difference and the separation distance, with the formula: \[ E = \frac{V}{d} \]
  • Due to this uniform field, equipotential surfaces within the capacitor appear as parallel planes.
This elegant design creates a predictable environment for studying electrostatic principles and is widely used in electronic circuits for its ability to hold and discharge electrical energy efficiently.
Electric Potential Difference
The electric potential difference, often referred to as voltage, is a measure of the work done to move a unit charge from one point to another in an electric field.
In the context of parallel plate capacitors, this difference is the driving force behind the electric field across the plates.
  • The potential difference is established when one plate gets a net positive charge while the opposite plate obtains a negative charge, creating an electric field between them.
  • In our scenario, this potential difference is set at 480 V, meaning each coulomb of charge experiences 480 joules of work as it moves across the plates.
The separations of this potential difference into smaller values along equipotential surfaces, like 120 V intervals, allow us to understand how potential energy changes spatially within the field.
Electric Field Lines
Electric field lines provide a visual representation of the direction and strength of an electric field.
In a parallel plate capacitor, these lines are drawn as straight and perpendicular from the positively charged plate to the negatively charged one.
  • The density of these lines indicates the field's strength; more lines mean a greater field intensity.
  • Field lines in capacitors are uniformly distributed, displaying a constant magnitude field between the plates.

The principle that electric field lines are perpendicular to equipotential surfaces holds true consistently. This perpendicular nature ensures that no work is needed to move a charge along an equipotential surface, maintaining energy conservation principles. When drawn correctly, these lines remind us of the fundamental idea that electric fields are vectors, possessing magnitude and direction.

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Most popular questions from this chapter

Two stationary point charges \(+3.00 \mathrm{nC}\) and \(+2.00 \mathrm{nC}\) are separated by a distance of 50.0 \(\mathrm{cm}\) . An electron is released from rest at a point midway between the two charges and moves along the line connecting the two charges. What is the speed of the electron when it is 10.0 \(\mathrm{cm}\) from the \(+3.00-\mathrm{charge}\) ?

Two charges of equal magnitude \(Q\) are beld a distance \(d\) apart. Consider only points on the line passing through both charges. (a) If the two charges have the same sign, find the location of all points (if there are any) at which (i) the potential (relative to infinity) is zero (is the electric field zero at these points?), and (ii) the electric field is zero (is the potential zero at these points?). and (b) Repeat part (a) for two charges having opposite signs.

(a) Show that \(V\) for a spherical shell of radius \(R\) , that has charge \(q\) distributed uniformly over its surface, is the same as \(V\) for a solid conductor with radius \(R\) and charge \(q\) . (b) You \(\mathrm{rub}\) an inflated balloon on the carpet and it acquires a potential that is 1560 \(\mathrm{V}\) lower than its potential before it became charged. If the charge is uniformly distributed over the surface of the balloon and if the radius of the balloon is \(15 \mathrm{cm},\) what is the net charge on the balloon? (c) In light of its \(1200-\mathrm{V}\) potential difference relative to you, do you think this balloon is dangerous? Explain.

Two very large, parallel metal plates carry charge densities of the same magnitude but opposite signs (Fig, 23.32\()\) . Assume they are close enough together to be treated as ideal infinite plates. Taking the potential to be zero at the left surface of the negative plate, sketch a graph of the potential as a function of \(x\) . Include all regions from the left of the plates to the right of the plates.

A point charge \(Q=+4.60 \mu \mathrm{C}\) is held fixed at the origin. A second point charge \(q=+1.20 \mu \mathrm{C}\) with mass of \(2.80 \times 10^{-4} \mathrm{kg}\) is placed on the \(x\) -axis, 0.250 \(\mathrm{m}\) from the origin. (a) What is the electric potential energy \(U\) of the pair of charges? (Take \(U\) to be zero when the charges have infinite scparation.) (b) The scond point charge is released from rest. What is its speed when its distance from the origin is (i) \(0.500 \mathrm{m} ;\) (ii) \(5.00 \mathrm{m} ;\) (iii) 50.0 \(\mathrm{m} ?\)
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