Electric field from a charged rod

In summary, the conversation covers the topic of finding the expression for the electric field at a point along the perpendicular bisector of a charged rod. The conversation also includes a discussion about the integral involved in the expression and a suggestion to change the origin to simplify the integral. The conversation ends with the correction that the formula was actually correct but still difficult to solve.
  • #1
MaximumTaco
45
0
Ok, I'm inexperienced with the latex code, so excuse me.

I'm trying to derive the expression for the electric field at a point, along the perpendicular bisector of a charged rod, of known total charge, length L, a perpendicular distance y from the rod.

The horizontal components cancel,
[tex]
\frac{qy}{4\pi \varepsilon_0 L} \int_0^L \frac{dx}{((\frac{L}{2}-x)^2 +y^2)^{3/2}}
[/tex]
I don't think this is right, because that seems a nasty integral. Any ideas?

Apologies if you think this is the wrong section. Thanks.
 
Last edited:
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  • #2
I finally got the tex perfect. Can somebody please help me as soon as possible? Thanks alot.
 
  • #3
The integral's not too bad.. Hint:
[tex]\int (1+x^2)^{-3/2}dx=x/\sqrt{1+x^2}[/tex]
 
  • #4
MaximumTaco said:
Ok, I'm inexperienced with the latex code, so excuse me.

I'm trying to derive the expression for the electric field at a point, along the perpendicular bisector of a charged rod, of known total charge, length L, a perpendicular distance y from the rod.

The horizontal components cancel,
[tex]
\frac{qy}{4\pi \varepsilon_0 L} \int_0^L \frac{dx}{((\frac{L}{2}-x)^2 +y^2)^{3/2}}
[/tex]
I don't think this is right, because that seems a nasty integral. Any ideas?

You are right, this is not right, and the integral is nasty.
You get a much nicer integral if you place your origin at the middle of the rod, and use the angle [tex]\varphi [/tex] instead of x. See pic.

[tex]r=\frac{y}{\cos(\varphi )} \mbox{ , } x = y \tan(\varphi )[/tex]...

ehild
 
Last edited:
  • #5
ehild said:
You are right, this is not right, and the integral is nasty.

ehild

Sorry, the formula was right... But it is still nasty.

ehild
 

1. What is an electric field?

An electric field is a physical field that surrounds any electrically charged object. It is a vector field, meaning it has both magnitude and direction, and it can exert a force on other charged objects within its range.

2. How is the electric field from a charged rod calculated?

The electric field from a charged rod can be calculated using Coulomb's Law, which states that the electric field strength is directly proportional to the magnitude of the charge on the rod and inversely proportional to the square of the distance from the rod.

3. What factors affect the strength of the electric field from a charged rod?

The strength of the electric field from a charged rod is affected by the magnitude of charge on the rod, the distance from the rod, and the medium in which the electric field is being measured. The electric field is stronger when the charge is larger and the distance is smaller, and it is weaker in materials with higher dielectric constants.

4. How does the direction of the electric field from a charged rod change with distance?

The direction of the electric field from a charged rod is always perpendicular to the rod at any given point. As the distance from the rod increases, the direction of the electric field becomes more radial, meaning it points away from the rod in all directions.

5. What are some real-world applications of the electric field from a charged rod?

The electric field from a charged rod has many practical applications, such as in electrostatic precipitators used to remove pollutants from smokestacks, in the operation of CRT televisions and computer monitors, and in particle accelerators used in scientific research. It also plays a crucial role in many electronics, such as capacitors and transistors.

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