# Potential difference and electric fields

1. Jul 5, 2009

### peaceandlove

1. The problem statement, all variables and given/known data
An electric field is given by E_x = (5 kN/C) x^3. Find the potential difference between the points on the x-axis at x= 1 m and x = 5 m. Answer in units of kV.

2. Relevant equations
V = -Ed

3. The attempt at a solution
V1 = -5*10^6 N/C * 1 = -5*10^6 volts
V2 = -5*10^6 * 5 = -25*10^6 volts = -2.5*10^7 volts

V2 - V1 = -25*10^6 -(-5*10^6) = -20*10^6 = -2.0*10^4 kV

I also tried cubing x and got -6.2e+5 but that is also wrong.

Last edited: Jul 5, 2009
2. Jul 5, 2009

### gabbagabbahey

$V=-Ed$ only when $E$ is uniform/constant along the length of $d$...is that the case here?

If not, you will need to use the more general equation:

$$V_b-V_a=-\int_a^b\vec{E}\cdot\vec{dl}$$

3. Jul 5, 2009

### peaceandlove

I would think that the x^3 would mean that E is not uniform/constant...

-5*10^6 times the integral of x^3 from 5 m to 1 m.... which is the same as -5*10^6*[(1/4)x^4] from 5 m to 1 m... which is equal to -7.8*10^5.

Is that it?

4. Jul 5, 2009

### gabbagabbahey

Where is the $10^6$ coming from?

5. Jul 5, 2009

### peaceandlove

I meant 10^4 since it's in kN/C, so the answer would be -7.8*10^3 kN?

Last edited: Jul 5, 2009
6. Jul 5, 2009

### peaceandlove

That apparently isn't the answer... what did I do wrong?

7. Jul 5, 2009

### diazona

$$1\,\mathrm{kN} = 10^3\,\mathrm{N}$$
and
$$1\,\mathrm{kV} = 10^3\,\mathrm{V}$$

Could it be that?

8. Jul 5, 2009

### peaceandlove

Oh wow... yet another stupid mistake. Thanks so much everyone!