# Calculating Electric Field from a Charged Filament

• tony873004
In summary, the problem involves finding the magnitude of the electric field at a distance of 0.67 mm from a long glass filament with a charge density of -2.7 nC/m. The correct formula to use is E = 2*k*lambda/r, where k is the Coulomb constant and lambda is the charge density. This formula can be derived through integration and Gauss's law, or it can be simplified from the formula used for a finite length filament.
tony873004
Gold Member

## Homework Statement

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?

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

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.

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?

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.

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.

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.

That works fine. Glad you got it.

## 1. What is an electric field?

An electric field is a physical field that surrounds a charged object and exerts a force on other charged objects within its vicinity. It is created by the presence of electric charges and is responsible for the movement of electrically charged particles.

## 2. How is the electric field of a filament calculated?

The electric field of a filament can be calculated using Coulomb's law, which states that the magnitude of the electric field at a given distance from a charged object is directly proportional to the magnitude of the charge and inversely proportional to the square of the distance between the two objects.

## 3. What factors affect the strength of the electric field of a filament?

The strength of the electric field of a filament is affected by the magnitude of the charge on the filament, the distance from the filament, and the permittivity of the medium surrounding the filament. The electric field is stronger when the charge is larger and weaker when the distance is greater or when the permittivity is lower.

## 4. How does the direction of the electric field of a filament relate to the direction of the electric current?

The direction of the electric field of a filament is always perpendicular to the direction of the electric current. This means that if the electric current is flowing in a straight line, the electric field will also be in a straight line, but at a right angle to the current.

## 5. Can the electric field of a filament be shielded or blocked?

Yes, the electric field of a filament can be shielded or blocked by insulating materials such as rubber or plastic. These materials do not conduct electricity and therefore prevent the electric field from reaching them.

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