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Chapter 23: Problem 29

A uniform electric field has magnitude \(E\) and is directed in the negative \(x\) -direction. The potential difference between point \(a\) (at \(x=0.60 \mathrm{m}\) ) and point \(b\) (at \(x=0.90 \mathrm{m} )\) is 240 \(\mathrm{V}\) . (a) Which point, \(a\) or \(b\) , is at the higher potential? (b) Calculate the value of E. (c) A negative point charge \(q=-0.200 \mu \mathrm{C}\) is moved from \(b\) to a. Calculate the work done on the point charge by the electric field.

Short Answer

Expert verified
a) Point a is at a higher potential. b) The electric field E is 800 V/m. c) The work done is -48 \(\mu J\).

Step by step solution

01

Understand Electric Potential

The electric potential decreases in the direction of the electric field. Since the electric field is directed in the negative x-direction, the potential decreases from point a to point b.
02

Determine Higher Potential Point

Since the electric potential decreases in the direction of the field, point a (at x = 0.60 m) is at a higher potential than point b (at x = 0.90 m).
03

Use Potential Difference Formula

The potential difference between two points is given by the formula: \( V = E \cdot d \), where \( d \) is the distance between the points. With a potential difference of 240 V and a distance of 0.90 m - 0.60 m = 0.30 m, we can find E.
04

Calculate E

Rearrange the potential difference formula to find \( E \): \( E = \frac{V}{d} = \frac{240 \text{ V}}{0.30 \text{ m}} = 800 \text{ V/m} \).
05

Calculate Work Done

The work done on a charge moved in an electric field is given by \( W = q \cdot V \), where \( q = -0.200 \mu C = -0.200 \times 10^{-6} C \) and \( V = 240 \text{ V} \). Hence, the work done is \( W = (-0.200 \times 10^{-6} C) \cdot 240 \text{ V} = -48 \times 10^{-6} J = -48 \mu J \).

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Key Concepts

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

Electric Potential
Electric potential is similar to gravitational potential energy. It is the potential energy stored due to an object's position within an electric field. Think of it as the energy that influences how a charged particle will move. When a test charge (positive or negative) is placed in an electric field, it has a certain potential energy depending on its location within that field.

The potential energy per unit charge is called electric potential, often denoted as "V" and measured in volts (V). Here is the formula to compute it:
  • Electric Potential (V) = Electric Potential Energy / Charge
Understanding this interaction helps predict how a charge will behave when it's released in the electric field. In scenarios where a uniform electric field is involved, realizing that the potential decreases in the direction of the field鈥檚 force gives insight into the forces acting upon the charge.
Potential Difference
Potential difference is the change in electric potential energy between two points in a field. It tells us how much energy per charge we will gain or lose when moved between two points. The difference is measured in volts and is often referred to as "voltage."

To find the potential difference, we often use the concept that the electric field will act along a path, causing the potential to change. In a uniform electric field, this can be calculated using:
  • Potential Difference (V) = Electric Field (E) x Distance (d)
In our exercise, the potential difference of 240 V indicates that point "a," being opposite the field direction, has a higher potential compared to point "b." Knowing the potential difference allows one to calculate the magnitude of the electric field, reinforcing the relationship that a greater difference will indicate a stronger field over a given distance.
Work Done in Electric Field
When a charge moves through an electric field, work is done on the charge or by the charge. This concept is crucial in understanding energy transfer within electric systems. The work done when moving a charge between two points is calculated via:
  • Work (W) = Charge (q) x Potential Difference (V)
In the context of our exercise, a negative charge is moved, and evaluating the work done involves simply multiplying the charge and the potential difference. This multiplication gives insight into how much energy is required to move that charge within the electric field.

Remember that if the work done is negative, as in our problem, it indicates that energy is being released by the charge while moving within the field's influence, as it naturally moves towards lower potential energy locations. Understanding this helps comprehend the energetic interactions and transitions in electrical contexts.

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Most popular questions from this chapter

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.

Two large, parallel, metal plates carry opposite charges of equal magnitude. They are separated by \(45.0 \mathrm{mm},\) and the potential difference between them is 360 \(\mathrm{V}\) . (a) What is the magnitude of the electric field (assumed to be uniform) in the region between the plates? (b) What is the magnitude of the force this field exerts on a particle with charge \(+2.40 \mathrm{nC} ?\) (c) Use the results of part (b) to compute the work done by the field on the particle as it moves from the higher-potential plate to the lower. (d) Compare the result of part (c) to the change of potential energy of the same charge, computed from the electric potential.

Three point charges, which initially are infinitely far apart, are placed at the corners of an equilateral triangle with sides \(d\) . Two of the point charges are identical and have charge \(q\) . If zerv net work is required to place the three charges at the comers of the triangle, what must the value of the third charge be?

A ring of diameter 8.00 \(\mathrm{cm}\) is fixed in place and carries a charge of \(+5.00 \mu \mathrm{C}\) unifurnly spread over its circumference. (a) How much work does it take to move a tiny \(+3.00-\mu \mathrm{C}\) charged ball of mass 1.50 \(\mathrm{g}\) 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 maxi- mum speed it will reach?

A small particle has charge \(-5.00 \mu \mathrm{C}\) and mass \(2.00 \times 10^{-4} \mathrm{kg}\) . It moves from point \(A\) , where the electric potential is \(V_{A}=+200 \mathrm{V},\) to point \(B\) , where the electric potential is \(V_{B}=+800 \mathrm{V}\) . The electric force is the only force acting on the particle. The particle has speed 5.00 \(\mathrm{m} / \mathrm{s}\) at point \(A .\) What is its speed at point \(B ?\) Is it moving faster or slower at \(B\) than at \(A ?\) Explain.
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