Determine electric field at x=1m

AI Thread Summary
To determine the electric field at x=1m, the electric potential function V=ax^2+bx+c is used, where a=13 V/m², b=-14 V/m, and c=50 V. The electric field E is the negative derivative of the potential V with respect to x, not simply V/r. At x=1m, the calculated potential is V=49 V, but this does not equal the electric field value. The correct method involves differentiating V to find E, which reveals that the electric field is not directly equal to the potential at that point. Additionally, the discussion clarifies that V=kQ/r is specific to point charges and not applicable here.
jaydnul
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Homework Statement


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

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

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

Homework Equations


∫Edx=V

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?
 
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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.
 
E=\frac{V}{r}. \frac{49}{1}=49. What am I doing wrong?
 
E=V/r is not right.
E is the derivative of V.
 
Oh wow. So V=k\frac{Q}{r} isn't correct?
 
Jd0g33 said:
Oh wow. So V=k\frac{Q}{r} isn't correct?

Not for this problem. V=k\frac{Q}{r} only applies for situations involving spherical charge symmetry, like a point charge.
 
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.
 
Perfect, thank you guys
 
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