Finding electric field via a given potential

1. Oct 4, 2011

DannyPhysika

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

The plastic rod of the length L=1 m has a non-uniform charge density λ=cx where positive constant c =2x10^-6 [some unit]. What unit c has to have? Find the electric potential at the point on the x axis 1 m to the left from the left end of the rod. Find the electric field at that point via potential. What is the direction of the field

2. Relevant equations

E = -dV/dx

3. The attempt at a solution

The first two parts I've solved. For the electric field:

I've gotten:

-3.47 x 10^3 N/C

or

-1.45 x 10^4 N/C

But I'm not sure which is good. Thanks for any information :)

2. Oct 4, 2011

ZenOne

Although I'm tempted to say that E will be equal to V because E is simply V/m and the question asks for the electric field 1m away, however, since the charge is non-uniform I assume you have to take the partial derivative of your derived expression which should be:

Ke*c([L-d*ln(1+(L/d))]

3. Oct 4, 2011

DannyPhysika

If I differentiate that with respect to d, I get that E = -(-3472) N/C which means that it is positive. But shouldn't it be negative?

4. Oct 4, 2011

SammyS

Staff Emeritus

5. Oct 4, 2011

DannyPhysika

I've done all my work on paper and I ended up with E(x) = -kc[ln(L-x) + x/(L-x) -1 -ln|x|] where x = -d where d is 1m. It's been a couple days but I plugged in my numbers and got 1.45 x 10^4 N/C. But this is a positive answer, shouldn't the electric field be to the left/negative?

6. Oct 4, 2011

SammyS

Staff Emeritus
Isn't the positively charged end of the rod nearest to x = 1m? Then E field is to the right.

7. Oct 4, 2011

SammyS

Staff Emeritus

4.225×103 V/m

8. Oct 5, 2011

DannyPhysika

I've just spent time with another student with this problem and we're really lost because we've got 4 different answers from different sources and not one of them are the same. We've gotten 8.99 x 10^3 N/C, -3.47 x 10^3, -1.45 x 10^4 N/C, and 4.225 x 10^4 N/C from the above poster. Can anyone else say that they've gotten one of the above answers? Thank you for any confirmation.

9. Oct 5, 2011

SammyS

Staff Emeritus
Well, I see that I've mis-read the problem. I had the rod centered at the origin, and found the E field at x = 1m .

I don't see any information about where the rod is located relative to the origin. I do see (now) that the question asks for the E field at a location 1 meter to the left of the end of the rod. Since linear charge density, λ, depends upon x, the rod's location relative to the origin is very important.

10. Oct 5, 2011

DannyPhysika

Yes it is a rather badly worded problem...

But yes, from my discussion with other students, we placed the rod's extreme left end at the origin. I wish I could show my work but I have it on paper and this is the last thing I have not done for this 5 problem assignment. I will have to pick one of the answers before class starts, I just wish I had more of a unanimous consensus on its value.

11. Oct 5, 2011

SammyS

Staff Emeritus
Well with that placement of the rod, the e field at x = -1 meter is definitely to the left (negative).

I'll look again, quickly.

What I have this time:

-1.738×103V/m

The total charge on the rod is c/2 Coulombs = 1.0×10-6 C, with 3/4 of that being to the right of x = 1/2.

The electric field at x = -1 m due to a point charge at x = 1/2 m, is E = -4×103 V/m.

That may help more than my rushed integration & differentiation & plugging stuff in.

Last edited: Oct 5, 2011
12. Oct 5, 2011

DannyPhysika

If V = kc (integral) (x)dx/(d+x)

then shouldn't E = -dV/dx where dV/dx just equals (d/dx)(kc (integral) (x)dx/(d+x)) which equals (kc(x))/(d+x)

so E = -(kc)(1/2) = -8.99 x 10^3 N/C ?

13. Oct 5, 2011

SammyS

Staff Emeritus
I different approach tells me I like the -3.47 ... answer .

14. Oct 5, 2011

SammyS

Staff Emeritus
Not no, but ... No Way.

For one thing in the integral x refers to a location along the rod, while in dV/dx, the x refers to the position at which you are finding the E field.

Also how did you let the x from the numerator jump out of the integral?

15. Oct 5, 2011

DannyPhysika

Ok I am also getting 3472.8 but I'm getting positive because it's:

E = -kc((x/x+L) - ln|x+L| + 1 + ln|x|) which gives -kc(-.19) which is 3472.8?

16. Oct 5, 2011

SammyS

Staff Emeritus
The field is definitely to the left if the rod lays from 0 to 1 on the x-axis.

17. Oct 6, 2011

ehild

The potential at r distance from a charge element dq is equal to kdq/r. If the rod of length L is placed with its left end in the origin, and the charge element is at distance $\xi$ from the origin, r=x-$\xi$ if x>L and r=$\xi$-x if x<0. See picture.

For negative x values then, the potential is

$U(x)=k\lambda \int_0^L{\frac{\xi}{\xi-x}d\xi}=k\lambda (L+x ln(1-\frac{L}{x}))$.

The electric field is shown in the second picture.

ehild

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Last edited: Oct 6, 2011