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Electric Fields/ Electric Potential

  • Thread starter Cilabitaon
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  • #1
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Homework Statement


Figure 1 shows a pair of parallel metal plates, A and B, fixed vertically 20mm apart with a potential difference of 1500V between them.
(a)(i) Draw the electric field lines in the space between the plates and calculate the electric field strength at point P.
(a)(ii)Sketch a graph showing the potential at different points in space between the plates.

Homework Equations


[tex]E_{f}=\frac{V}{x}[/tex]


The Attempt at a Solution


The first part is shown in the diagram by my (rather scruffy)red lines; they are drawn better on my actual diagram. :redface:

For the potential at P:

[tex]E_{P}=\frac{1500/2}{(20 \times 10^{-3})/2} = 7.5 \times 10^{4}Vm^{-1}[/tex]

For the last part I have a 100mmx60mm piece of graph paper with the labels 'potential' and 'distance' on the vertical and horizontal axes respectively.

Now, in my thinking all I can get from this is that the graph must start at 0, and the gradient must be [tex]E[/tex].

c.f. [tex]y = mx + c[/tex] and you get [tex]V = Ex (+ 0)[/tex].

I have my values for [tex]V[/tex] from (0>1500)V on my potential axis and my values for [tex]x[/tex] from (0>20.0)e-3m.

The problem with this is I can see everything that is going on, but I just cannot draw a graph of it; and it's really starting to annoy, so any help would be appreciated.
 

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Answers and Replies

  • #2
Cyosis
Homework Helper
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You can draw the graph in the same way you draw any linear function. You know it is a straight line so you only need two points. However you made a mistake in calculating the electric field. You are using the potential difference of 1500V at the point P, but the potential difference between plate A and P is not 1500.
 
  • #3
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You can draw the graph in the same way you draw any linear function. You know it is a straight line so you only need two points. However you made a mistake in calculating the electric field. You are using the potential difference of 1500V at the point P, but the potential difference between plate A and P is not 1500.
Right, it's half way between the two plates in a uniform field so the potential drop is half...right?

So this would mean it's just [tex]E = \frac{1500/2}{(20 \times 10^{-3})/2} = 7.5 \times 10 ^{4}Vm^{-1}[/tex]
 
  • #4
Cyosis
Homework Helper
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Yes that is correct. You know that the eletric field does not change between the plates so calculating the field at any point gives you the field for all points (within the capacitor).
 
  • #5
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It's quite worrying to me that I had such problems with a [tex]V \propto x[/tex] relationship :rofl:
 

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