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Learning objectives

By the end of this section, you will be able to:

  • Describe the relationship between voltage and electric field.
  • Derive an expression for the electric potential and electric field.
  • Calculate electric field strength given distance and voltage.

The information presented in this section supports the following AP® learning objectives and science practices:

  • 2.C.5.2 The student is able to calculate the magnitude and determine the direction of the electric field between two electrically charged parallel plates, given the charge of each plate, or the electric potential difference and plate separation. (S.P. 2.2)
  • 2.C.5.3 The student is able to represent the motion of an electrically charged particle in the uniform field between two oppositely charged plates and express the connection of this motion to projectile motion of an object with mass in the Earth’s gravitational field. (S.P. 1.1, 2.2, 7.1)
  • 2.E.3.1 The student is able to apply mathematical routines to calculate the average value of the magnitude of the electric field in a region from a description of the electric potential in that region using the displacement along the line on which the difference in potential is evaluated. (S.P. 2.2)
  • 2.E.3.2 The student is able to apply the concept of the isoline representation of electric potential for a given electric charge distribution to predict the average value of the electric field in the region. (S.P. 1.4, 6.4)

In the previous section, we explored the relationship between voltage and energy. In this section, we will explore the relationship between voltage and electric field. For example, a uniform electric field E size 12{E} {} is produced by placing a potential difference (or voltage) Δ V size 12{V} {} across two parallel metal plates, labeled A and B. (See [link] .) Examining this will tell us what voltage is needed to produce a certain electric field strength; it will also reveal a more fundamental relationship between electric potential and electric field. From a physicist’s point of view, either Δ V size 12{V} {} or E size 12{E} {} can be used to describe any charge distribution. Δ V size 12{V} {} is most closely tied to energy, whereas E size 12{E} {} is most closely related to force. Δ V size 12{V} {} is a scalar    quantity and has no direction, while E size 12{E} {} is a vector    quantity, having both magnitude and direction. (Note that the magnitude of the electric field strength, a scalar quantity, is represented by E size 12{V} {} below.) The relationship between Δ V size 12{V} {} and E size 12{E} {} is revealed by calculating the work done by the force in moving a charge from point A to point B. But, as noted in Electric Potential Energy: Potential Difference , this is complex for arbitrary charge distributions, requiring calculus. We therefore look at a uniform electric field as an interesting special case.

The figure shows two vertically oriented parallel plates A and B separated by a distance d. The plate A is positively charged and B is negatively charged. Electric field lines are parallel between the plates and curved at the ends of the plates. A charge q is moved from A to B. The work done W equals q times V sub A B, and the electric field intensity E equals V sub A B over d and potential difference delta V equals q times V sub A B.
The relationship between V size 12{V} {} and E size 12{E} {} for parallel conducting plates is E = V / d size 12{E=V/d} {} . (Note that Δ V = V AB size 12{ΔV=V rSub { size 8{"AB"} } } {} in magnitude. For a charge that is moved from plate A at higher potential to plate B at lower potential, a minus sign needs to be included as follows: –Δ V = V A V B = V AB . See the text for details.)

The work done by the electric field in [link] to move a positive charge q size 12{q} {} from A, the positive plate, higher potential, to B, the negative plate, lower potential, is

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
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A mouse of mass 200 g falls 100 m down a vertical mine shaft and lands at the bottom with a speed of 8.0 m/s. During its fall, how much work is done on the mouse by air resistance
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emma
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Krampah Reply
2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
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Samuel Reply
can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
Joseph Reply
Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
Joseph
"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
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Mujahid
A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
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Source:  OpenStax, College physics for ap® courses. OpenStax CNX. Nov 04, 2016 Download for free at https://legacy.cnx.org/content/col11844/1.14
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