# Electric field of a filament

1. Feb 10, 2008

### tony873004

1. The problem statement, all variables and given/known data
A long glass filament carries a charge density $$\lambda = - 2.7{\rm{ nC/m}}$$ . What is the magnitude of the electric field 0.67 mm from the filament?
3. The attempt at a solution

Did I do this problem correctly? I'm guessing not because my answer has an extra /m . Shouldn't it simply be N/C?
$$E = \left| {\frac{{k\,d\lambda }}{{r^2 }}} \right| = \left| {\frac{{8.99 \times 10^9 {\rm{N}} \cdot \frac{{{\rm{m}}^{\rm{2}} }}{{{\rm{C}}^{\rm{2}} }} \cdot - 2.7{\rm{ nC/m}} \cdot 10^{ - 9} {\rm{C/nC}}}}{{{\rm{0}}{\rm{.67mm}} \cdot 0.001{\rm{m/mm}}}}} \right| = 54000000{\rm{ N/Cm}}$$

Last edited: Feb 10, 2008
2. Feb 10, 2008

### awvvu

Isn't the filament a long line of charge? So, the field strength is proportional to 1/r.

3. Feb 10, 2008

### Dick

Basically, except isn't E=2*k*lambda/r? I think you missed a '2'. Your equation is for the infinitesimal element of E before it is integrated.

4. Feb 10, 2008

### tony873004

Thanks. Ok, then would it be:
$$E = \left| {\frac{{2k\lambda }}{r}} \right| = \left| {\frac{{2 \cdot 8.99 \times 10^9 {\rm{N}} \cdot \frac{{{\rm{m}}^{\rm{2}} }}{{{\rm{C}}^{\rm{2}} }}\left( { - 2.7{\rm{nC/m}} \cdot 10^{ - 9} {\rm{C/nC}}} \right)}}{{{\rm{0}}{\rm{.67mm}} \cdot 0.001{\rm{m/mm}}}}} \right| = 72000\,{\rm{N/C}}$$
?
Where did E=2k*lambda/r come from? Is this just a standard formula I can use as a starting point, or would I be expected to integrate and come up with this formula?

5. Feb 10, 2008

### awvvu

6. Feb 10, 2008

### Dick

I know it's k*lambda/r. I had to look it up to get the numerical factor of 2. Just looking at the problem statement, it doesn't look like a 'derive the formula' type problem, though the derivation isn't that hard. I would suspect the formula is in your book or notes and you can just use it. Check though.

7. Feb 10, 2008

### tony873004

Thanks. I'd like to understand how to do it even if we can start with the formula. Awvvu's link shows how. Stay tuned... I might have questions about that link.

8. Feb 10, 2008

### tony873004

I though of another way. From our class notes, we did the integration of the same problem, except the filament had a finite length. The formula was $${\overrightarrow {E_y } = \frac{{k\lambda L}}{{y^2 \sqrt {1 + \left( {\frac{L}{{2y}}} \right)^2 } }}}$$

To convert this to use for an infinite length:

As L approaches infinity, the 1 under the radical becomes insignificant, allowing me to eliminate it and replace $$\sqrt {\left( {\frac{L}{{2y}}} \right)^2 }$$ with $${\frac{L}{{2y}}}$$.
. Then my L's cancel leaving me with $$\frac{{2k\lambda }}{y}$$

Thanks for checking my work. I probably would have submitted my original attempt if the two of you didn't correct me.

9. Feb 10, 2008

### Dick

That works fine. Glad you got it.