Finding electric field via a given potential

  • #1

Homework Statement



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


Homework Equations



E = -dV/dx

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 :)
 

Answers and Replies

  • #2
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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
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))]

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
SammyS
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Show how you got your answers so we can help you.
 
  • #5
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
SammyS
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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?

Isn't the positively charged end of the rod nearest to x = 1m? Then E field is to the right.
 
  • #7
SammyS
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I got somewhat different answers.

4.225×103 V/m
 
  • #8
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
SammyS
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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.

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
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
SammyS
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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.

Added in Edit:

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:
  • #12
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
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I different approach tells me I like the -3.47 ... answer .
 
  • #14
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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 ?

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
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
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The field is definitely to the left if the rod lays from 0 to 1 on the x-axis.
 
  • #17
ehild
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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 [itex]\xi[/itex] from the origin, r=x-[itex]\xi[/itex] if x>L and r=[itex]\xi[/itex]-x if x<0. See picture.

For negative x values then, the potential is

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

The electric field is shown in the second picture.

ehild
 

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