# Determine electric field at x=1m

1. Sep 16, 2013

### jaydnul

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
$∫Edx=V$

3. 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?

2. Sep 16, 2013

### Staff: Mentor

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. Sep 16, 2013

### jaydnul

$E=\frac{V}{r}$. $\frac{49}{1}=49$. What am I doing wrong?

4. Sep 16, 2013

### Staff: Mentor

E=V/r is not right.
E is the derivative of V.

5. Sep 16, 2013

### jaydnul

Oh wow. So $V=k\frac{Q}{r}$ isn't correct?

6. Sep 16, 2013

### collinsmark

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

7. Sep 16, 2013

### collinsmark

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. Sep 16, 2013

### jaydnul

Perfect, thank you guys