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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.

Short Answer

Expert verified
(a) 8000 V/m, (b) 1.92 \times 10^{-5} N, (c) 8.64 \times 10^{-7} J, (d) Work equals change in potential energy.

Step by step solution

01

Determine the Electric Field (a)

To find the electric field (assumed uniform) between the plates, we use the formula for the electric field due to a potential difference: \( E = \frac{V}{d} \), where \( V = 360 \text{ V} \) and \( d = 45.0 \text{ mm} = 0.045 \text{ m} \). Substitute the values: \( E = \frac{360}{0.045} \), which gives \( E = 8000 \text{ V/m} \). The magnitude of the electric field is 8000 \(\text{ V/m}\).
02

Calculate the Force on a Charge (b)

The force exerted on a charge in an electric field is calculated using \( F = qE \). Here \( q = 2.40 \text{ nC} = 2.40 \times 10^{-9} \text{ C} \) and \( E = 8000 \text{ V/m} \). Thus, \( F = 2.40 \times 10^{-9} \times 8000 \). This results in \( F = 1.92 \times 10^{-5} \text{ N} \). The magnitude of the force exerted on the particle is \(1.92 \times 10^{-5} \text{ N}\).
03

Compute Work Done by the Field (c)

The work done by the electric field is given by \( W = q \Delta V \). We already know \( q = 2.40 \times 10^{-9} \text{ C} \) and \( \Delta V = 360 \text{ V} \). Thus, \( W = 2.40 \times 10^{-9} \times 360 \), which gives \( W = 8.64 \times 10^{-7} \text{ J} \). The work done by the field on the particle is \(8.64 \times 10^{-7} \text{ J}\).
04

Compare Work to Change in Potential Energy (d)

The change in electric potential energy for a charge moving in an electric field is given by \( \Delta U = q \Delta V \). Since \( W = q \Delta V \) and work-energy principle states that work done by the field is equal to the change in potential energy, \( \Delta U = W \). Thus, \( \Delta U = 8.64 \times 10^{-7} \text{ J} \). The work done and the change in potential energy are the same, confirming our calculations.

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Key Concepts

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

Potential Difference
The concept of potential difference, often referred to as voltage, is central to understanding electric fields. It is defined as the amount of work needed to move a charge between two points in an electric field.
In the context of two parallel plates, the potential difference between the plates is the driving factor for the electric field that is established.
  • The potential difference is measured in volts (V), and provides a measure of the energy difference per unit charge between the two points.
  • In this scenario, the potential difference of 360 V is distributed uniformly across the separation of 45 mm between the plates.
  • This uniform distribution allows us to assume that the electric field between the plates is uniform too.
Understanding potential difference helps us to calculate the electric field strength, as described by the formula:
\[ E = \frac{V}{d} \]
This simple formula is essential for determining how strong the electric field is and is crucial when trying to calculate the force on a charge placed in such a field.
Force on Charge
The force on a charge within an electric field can be determined using the relationship between the charge, the electric field, and the force exerted.
In a uniform electric field, such as the one between the two metal plates, this force can be calculated by the formula:
\[ F = qE \]
where:
  • \( F \) is the force in newtons (N)
  • \( q \) is the charge in coulombs (C)
  • \( E \) is the electric field strength in volts per meter (V/m)
For a charge of +2.40 nC (nanocoulombs) in a field of 8000 V/m, this results in a force of:
  • \( F = 2.40 \times 10^{-9} \times 8000 \)
  • This gives a force of \( F = 1.92 \times 10^{-5} \text{ N} \)
The significance of knowing the force on a charge is that it tells us how the electric field affects charged particles, which is fundamental when analyzing electric circuits and fields.
Work Done by Electric Field
Work done by an electric field refers to the energy transfer that occurs when a charge moves within the field.
The work done by the field can be calculated using the formula:
\[ W = q \Delta V \]
where:
  • \( W \) is the work in joules (J)
  • \( q \) is the charge in coulombs (C)
  • \( \Delta V \) is the potential difference in volts (V)
In our specific example, the work done is:
  • With a charge of +2.40 nC and a potential difference of 360 V, we have: \( W = 2.40 \times 10^{-9} \times 360 \)
  • This results in \( W = 8.64 \times 10^{-7} \text{ J} \)
Understanding this concept is crucial because it connects the physical movement of charges to the potential energy changes within the field. This relationship is a fundamental aspect of electrical physics, explaining how energy is harvested and utilized in electric fields.
Change in Electric Potential Energy
The electric potential energy of a system refers to the potential an object has to do work due to its position within an electric field.
When a charge moves in an electric field, its electric potential energy changes, and is quantified by:\[ \Delta U = q \Delta V \]
This change in potential energy equals the work done by the electric field, thus:
  • The change in potential energy, \( \Delta U \)
  • Is the same as the work done, \( W = 8.64 \times 10^{-7} \text{ J} \)
This means that as the charge moves from a higher potential to a lower potential, the field performs work on the charge, showing a direct conversion of potential energy into work.
In practical terms, calculating the change in electric potential energy helps in understanding how electric fields store and transfer energy, and is key to the study of electricity and circuits.

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Most popular questions from this chapter

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.

A point charge has a charge of \(2.50 \times 10^{-11} \mathrm{C}\) . At what distance from the point charge is the electric potential (a) 90.0 \(\mathrm{V}\) and (b) 30.0 \(\mathrm{V}\) ? Take the potential to be zero at an infinite distance from the charge.

Alpha particles \(\left(\text { mass }=6.7 \times 10^{-27} \mathrm{kg}, \text { charge }=+2 e\right)\) are shot directly at a gold foil target. We can model the gold nucleus as a uniform sphere of charge and assume that the gold does not move. (a) If the radius of the gold nucleus is \(5.6 \times\) \(10^{-15} \mathrm{m}\) , what minimum speed do the alpha particles need when they are far away to reach the surface of the gold nucleus? (lanore relativistic effects.) (b) Give good physical reasons why we can ignore the effects of the orbital electrons when the alpha particle is (i) outside the electron orbits and (ii) inside the electron orbits.

A charge of 28.0 \(\mathrm{nC}\) is placed in a uniform electric field that is directed vertically upward and has a magnitude of \(4.00 \times 10^{4} \mathrm{V} / \mathrm{m}\) . What work is done by the electric force when the charge moves (a) 0.450 \(\mathrm{m}\) to the right; (b) 0.670 \(\mathrm{m}\) upward; (c) 2.60 \(\mathrm{m}\) at an angle of \(45.0^{\circ}\) downward from the horizontal?

A thin insulating rod is bent into a semicircular are of radius \(a,\) and a total electric charge \(Q\) is distributed uniformly along the rod. Calculate the potential at the center of curvature of the are if the potential is assumed to be zero at infinity.

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