Determine electric field at x=1m

  • Thread starter jaydnul
  • Start date
  • #1
516
11

Homework Statement


A. The electric potential in a certain region is
[tex]V=ax^2+bx+c[/tex]
where [itex]a=13\frac{V}{m^2}[/itex], [itex]b=-14\frac{V}{m}[/itex], and [itex]c=50V[/itex].

Determine the electric field at [itex]x=1m[/itex].
Answer in units of [itex]\frac{V}{m}[/itex].

B. Determine the position where the electric field is zero. Answer in units of m.

Homework Equations


[itex]∫Edx=V[/itex]

The Attempt at a Solution


Well V=49. I have no doubt about that. So if V is 49 at x=1, wouldn't the electric field also be 49 at x=1?
 

Answers and Replies

  • #2
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There is no reason why the electric field should have the same numerical value as the potential.
Note that this comparison has no physical meaning anyway - potential and field have different units.
 
  • #3
516
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[itex]E=\frac{V}{r}[/itex]. [itex]\frac{49}{1}=49[/itex]. What am I doing wrong?
 
  • #4
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E=V/r is not right.
E is the derivative of V.
 
  • #5
516
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Oh wow. So [itex]V=k\frac{Q}{r}[/itex] isn't correct?
 
  • #6
collinsmark
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Oh wow. So [itex]V=k\frac{Q}{r}[/itex] isn't correct?
Not for this problem. [itex]V=k\frac{Q}{r}[/itex] only applies for situations involving spherical charge symmetry, like a point charge.
 
  • #7
collinsmark
Homework Helper
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Oh, and don't forget that there is a negative sign involved in the relationship between electric potential and electric field. The equation that you listed in the "Relevant Equations" section erroneously neglects this.
 
  • #8
516
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Perfect, thank you guys
 

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