025 Query 25

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course PHY 202

025. Query 25 was submitted 15 July 2011 around 12:15 PM.

025. Query 25

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Question: `qQuery introductory set #1, 10-17

Explain how to find the potential difference in volts between two given points on the x axis, due to a given charge at the origin.

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Your solution:

Potential difference is the work per Coulomb of charge moved between the two points:

potential difference is dV = V2 - V1.

To find this work multiply the average force on a Coulomb of charge by the displacement from the first point to the second: the work required to move the charge is W = Q `dV.

The approximate average force can be determined by finding the force on a 1 Coulomb test charge at the two points and averaging the two forces. Multiplying this average force by the displacement gives an approximate potential difference.

When a charge Q moves from a point at potential V1 to a point at potential V2, it passes through a potential difference `dV = V2 - V1. The potential difference is measured in volts = Joules / Coulomb; the charge is measured in Coulombs.

• Moving a positive charge through a positive potential difference is analogous to raising a mass against the force of gravity, which requires positive work.

• So the work done to get the charge through the potential difference is W = Q `dV.

• The total work required is found by adding up the contributions from all the subintervals.

• A Volt is a measure of work per unit charge, measured in Joules per Coulomb.

confidence rating #$&*:

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Given Solution:

`a** Potential difference is the work per Coulomb of charge moved between the two points. To find this work you can multiply the average force on a Coulomb of charge by the displacement from the first point to the second.

You can find an approximate average force by finding the force on a 1 Coulomb test charge at the two points and averaging the two forces. Multiplying this average force by the displacement gives an approximate potential difference.

Since the force is not a linear function of distance from the given charge, if the ratio of the two distances from the test charge is not small the approximation won't be particularly good. The approximation can be improved to any desired level of accuracy by partitioning the displacement between charges into smaller intervals of displacement and calculating the work done over each. The total work required is found by adding up the contributions from all the subintervals.

University Physics students should understand how this process yields the exact work, which is the integral of the force function F(x) = k Q / x^2 between the two x values, yielding total work W = k * Q * 1 Coulomb (1 / x1 - 1 / x2) and potential difference V = k * Q (1 / x1 - 1 / x2). **

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Self-critique (if necessary): OK

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Self-critique Rating: 3

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Question: `qExplain how to find the potential difference between two points given the magnitude and direction of the uniform electric field between those points.

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Your solution:

E = dV / dx is called the electric field. So the work per Coulomb done between the two points is equal to the product of the electric field E and the displacement `dr. Thus for constant field E we have V = E * `dr.

confidence rating #$&*:

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Given Solution:

`a** The work per Coulomb done between the two points is equal to the product of the electric field E and the displacement `dr. Thus for constant field E we have V = E * `dr. **

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Self-critique (if necessary): OK

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Self-critique Rating: 3

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Question: `qExplain how to find the average electric field between two points given a specific charge and the work done on the charge by the electric field as the charge moves between the points.

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Your solution:

The potential difference is the product of the potential gradient and the displacement: `dV = dV / dx * `ds.

E = dV / dx is called the electric field. Since E = Fave / q, we see that the electric field E is the force per unit charge experienced by a charge.

The average electric field is Eave = `dV / `ds

confidence rating #$&*:

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Given Solution:

`a** You get ave force from work and distance: F_ave = `dW / `ds.

You get ave electric field from work and charge: E_ave = F / q.

An alternative:

Find potential difference `dV = `dW / q.

Ave electric field is Eave = `dV / `ds **

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Self-critique (if necessary): OK

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Self-critique Rating: 3

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Question: `qIn your own words explain the meaning of voltage.

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Your solution:

Voltage is a representation of the electric potential energy per unit charge measured in Joules per Coulomb.

If a unit of electrical charge were placed in a location, the voltage indicates the potential energy of it at that point. In other words, it is a measurement of the energy contained within an electric field, or an electric circuit, at a given point. V = E*d

confidence rating #$&*:

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Given Solution:

`a** Voltage is the work done per unit of charge in moving charge from one point to another. **

STUDENT SOLUTION

Voltage is the difference in electric field times the distance between two points:

V = E*d

The bigger the difference in voltage between two points, the greater potential to drive electrical current.

INSTRUCTOR COMMENT

Your answer was also correct.

Note that the units for voltage are Joules / Coulomb, which is consistent with the given solution (work per unit of charge).

However this unit can also be expressed as N * m / C , or (N / C) * m, consistent with your statement (N / C is the unit of electric field, so this would be the product of electric field and distance).

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Self-critique (if necessary): OK

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Self-critique Rating: 3

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