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Chapter 23: Problem 42

A very large plastic sheet carries a uniform charge density of \(-\)6.00 nC\(/\)m\(^2\) on one face. (a) As you move away from the sheet along a line perpendicular to it, does the potential increase or decrease? How do you know, without doing any calculations? Does your answer depend on where you choose the reference point for potential? (b) Find the spacing between equipotential surfaces that differ from each other by 1.00 V. What type of surfaces are these?

Short Answer

Expert verified
(a) The potential increases as you move away. (b) The spacing is 2.94 mm; these are parallel planes.

Step by step solution

01

Understanding the Electric Field Direction

For a large uniformly charged sheet, the electric field is uniform and perpendicular to the surface. Since the sheet has a negative charge density, the electric field direction is pointing towards the sheet. When moving away from the sheet against the direction of the field, the electric potential increases.
02

Potential Change as a Conceptual Understanding

The potential decreases as you move with the electric field and increases as you move against it. Because you are moving against the electric field direction (moving away from a negatively charged sheet), the potential increases. The potential consistent change does not depend on the reference point.
03

Electric Field Magnitude for a Charged Sheet

The electric field (E) of an infinite charged sheet with charge density \( \sigma \) is given by \( E = \frac{\sigma}{2\varepsilon_0} \) where \( \varepsilon_0 \) is the permittivity of free space.
04

Calculate the Electric Field Strength

Given \( \sigma = -6.00 \, \text{nC/m}^2 = -6.00 \times 10^{-9} \, \text{C/m}^2 \), the electric field magnitude is \[ E = \frac{-6.00 \times 10^{-9}}{2 \times 8.85 \times 10^{-12}} \, \text{N/C} = -339.83 \, \text{N/C}. \]
05

Determine Equipotential Surface Spacing

Equipotential surfaces are perpendicular to the electric field lines. The change in potential \( \Delta V = E \cdot d \) where \( d \) is the spacing between equipotential surfaces. Solving for \( d \) gives \[ d = \frac{\Delta V}{E} = \frac{1.00 \text{ V}}{339.83 \, \text{N/C}} \approx 0.00294 \, \text{m}\approx 2.94 \, \text{mm}. \]
06

Understand the Type of Equipotential Surfaces

The equipotential surfaces related to a large plane of uniform charge are planar and parallel surfaces to the sheet itself, indicating consistent spacing between them.

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Key Concepts

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

Electric Potential
Electric potential, or simply potential, is a crucial concept in understanding electric fields and forces. It鈥檚 comparable to gravitational potential energy, but in this case, it鈥檚 related to electric charges instead of masses. Electric potential signifies the work done to move a unit charge from one point to another in an electric field.

When discussing electric potential in relation to charged sheets, it鈥檚 important to remember that potential always increases when moving against an electric field direction; it decreases when moving with it. As the exercise points out, when moving away from a negatively charged sheet you increase the potential because you are moving against the field direction. This change occurs regardless of where you choose your reference point, as the potential difference is what truly matters.
  • Gravitational Analogy: Just as climbing a hill increases gravitational potential energy, moving toward a like charge increases electric potential energy.

  • Reference Point Independence: The key is not the absolute potential, but rather its change; the choice of zero potential is arbitrary.
Equipotential Surfaces
Equipotential surfaces are an essential concept that aids in visualizing electric fields. An equipotential surface consists of points that all have the same electric potential. This means that no work is needed to move a charge along the surface.

In the case of an infinite charged sheet, these equipotential surfaces are planes parallel to the sheet. Since they are orientated perpendicular to the electric field lines from the charged sheet, they reflect the uniform nature of the electric field.
  • Uniformity: For a uniformly charged sheet, equipotential surfaces are consistently spaced out. This provides a clear indication of the electric field鈥檚 strength.

  • Energy Efficiency: Moving along these surfaces doesn鈥檛 change the electric potential energy, meaning any movement perpendicular to these surfaces results in a change in potential, mirroring the work done against or with the electric field.
Charge Density
Charge density is another pivotal concept when addressing electric fields and potentials. It refers to the amount of electric charge per unit area for a surface charge density, represented as \( \sigma \). In the given problem, the charge density is negative, which has implications for the direction and nature of the electric field.

A negative charge density indicates that the electric field lines are directed towards the charged sheet, rather than away. This inversion leads to the potential increasing as one moves away from the sheet, exemplifying how charge density affects electric potential and field direction.
  • Understanding \( \sigma \): It characterizes the electric field strength and orientation near the surface, being crucial for calculations involving electric field and potential.

  • Critical Inversion: A negative \( \sigma \) ensures electric fields point towards the charge source, influencing how potential changes with distance.

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Most popular questions from this chapter

A ring of diameter 8.00 cm is fixed in place and carries a charge of \(+\)5.00 \(\mu\)C uniformly spread over its circumference. (a) How much work does it take to move a tiny \(+\)3.00-\(\mu\)C charged ball of mass 1.50 g from very far away to the center of the ring? (b) Is it necessary to take a path along the axis of the ring? Why? (c) If the ball is slightly displaced from the center of the ring, what will it do and what is the maximum speed it will reach?

A solid conducting sphere has net positive charge and radius \(R =\) 0.400 m. At a point 1.20 m from the center of the sphere, the electric potential due to the charge on the sphere is 24.0 V. Assume that \(V = 0\) at an infinite distance from the sphere. What is the electric potential at the center of the sphere?

For a particular experiment, helium ions are to be given a kinetic energy of 3.0 MeV. What should the voltage at the center of the accelerator be, assuming that the ions start essentially at rest? (a) -3.0 MV; (b) +3.0 MV; (c) +1.5 MV; (d) +1.0 MV.

A positive point charge \(q_1 = +5.00 \times 10^{-4}\) C is held at a fixed position. A small object with mass 4.00 \(\times 10^{-3}\) kg and charge \(q_2 = -3.00 \times 10^{-4}\) C is projected directly at \(q_1\) . Ignore gravity. When \(q_2\) is 0.400 m away, its speed is 800 m\(/\)s. What is its speed when it is 0.200 m from \(q_1\) ?

The vertical deflecting plates of a typical classroom oscilloscope are a pair of parallel square metal plates carrying equal but opposite charges. Typical dimensions are about 3.0 cm on a side, with a separation of about 5.0 mm. The potential difference between the plates is 25.0 V. The plates are close enough that we can ignore fringing at the ends. Under these conditions: (a) how much charge is on each plate, and (b) how strong is the electric field between the plates? (c) If an electron is ejected at rest from the negative plate, how fast is it moving when it reaches the positive plate?
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