Electric Field due to line charge

In summary, the problem involves a non-conducting rod with a uniformly distributed positive charge of linear charge density, inclined at an angle of 30° with the +ve Y-axis. The electrostatic field at the origin due to the rod is E_0 N/C and its direction is along line OC. The value of θ is given as 10a+b degrees and the goal is to determine the value of (a+b). The solution involves finding the expressions for the x and y components of the field, integrating over the line, and using trigonometry to find the value of θ. Since this is an integer type question, a and b are integers that can be determined by trial and error.
  • #1
utkarshakash
Gold Member
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Homework Statement


A non-conducting rod AB, having uniformly distributed positive charge of linear charge density λ is kept in x-y plane. The rod AB is inclined at an angle 30° with +ve Y-axis. The magnitude of electrostatic field at origin due to rod AB is E_0 N/C and its direction is along line OC. If line OC makes an angle θ=10a+b degree with negative x-axis as shown in the figure, calculate the value of (a+b) [OA=2m and λ=10^3 C/m]

Homework Equations


Please see the attached diagram

The Attempt at a Solution



[itex]E_x = \dfrac{\lambda}{4 \pi \epsilon _0 d} (\sin 60 + \sin 30) \\
E_y = \dfrac{\lambda}{4 \pi \epsilon _0 d} |\cos 30 - \cos 60| \\

\tan \alpha = \dfrac{E_y}{E_x} \\
=\dfrac{ |\cos 30 - \cos 60| }{(\sin 60 + \sin 30)} [/itex]

which comes out to be 15°.

Thus, θ = 30° - 15° = 15°. So, a+b = 6. But it's not the correct answer :(
 

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  • #2
I had a different result for ##E_y## but the same result for ##E_x##. How did you arrive at your expressions of the field components? I integrated over the line.
 
  • #3
guitarphysics said:
I had a different result for ##E_y## but the same result for ##E_x##. How did you arrive at your expressions of the field components? I integrated over the line.

I did the integration again and got the expression for E_y as
[itex] \dfrac{\lambda}{16 \pi \epsilon _0 d} (\cos 60^0 - \cos 120^0) [/itex]

Do you get the same result?
 
  • #4
utkarshakash said:
A non-conducting rod AB, having uniformly distributed positive charge of linear charge density λ is kept in x-y plane. The rod AB is inclined at an angle 30° with +ve Y-axis. The magnitude of electrostatic field at origin due to rod AB is E_0 N/C and its direction is along line OC. If line OC makes an angle θ=10a+b degree with negative x-axis as shown in the figure, calculate the value of (a+b) [OA=2m and λ=10^3 C/m]

Hi utkarshakash

1. Is θ=10a+b degree given and value of a+b is to be determined ?
2. What are a and b ?
3. What is the answer according to the key ?

Thanks
 
Last edited:
  • #5
Vibhor said:
Hi utkarshakash

1. Is θ=10a+b degree given and value of a+b is to be determined ?
2. What are a and b ?
3. What is the answer according to the key ?

Thanks

1. Yes
2. a and b are variables which need to be determined.
3. I don't remember. I posted this a long time ago. I don't have the answer keys right now.
 
  • #6
utkarshakash said:
1. Yes
2. a and b are variables which need to be determined.
3. I don't remember. I posted this a long time ago. I don't have the answer keys right now.

Thank you :)

But I wonder how is this possible . Value of θ can be determined which gives value of (10a+b) ,but then we need another equation in a and b to determine value of (a+b)

Any thoughts ?
 
  • #7
Vibhor said:
Thank you :)

But I wonder how is this possible . Value of θ can be determined which gives value of (10a+b) ,but then we need another equation in a and b to determine value of (a+b)

Any thoughts ?

This is an integer type question. This means you'll have to get a and b by trial and error and I don't think it'd be difficult to do so.
 
  • #8
Fine . So a and b are integers . This is what I wanted to know .

Thanks a lot :)
 

FAQ: Electric Field due to line charge

What is an "Electric Field due to line charge"?

An "Electric Field due to line charge" is a physical concept used to describe the force experienced by charged particles in the presence of a line of charge. It is a vector quantity that represents the direction and magnitude of the force that a positive test charge would experience when placed at a given point in space near a line of charge.

How is the Electric Field due to line charge calculated?

The Electric Field due to line charge is calculated using the formula E = (λ/2𝜋ε0r) where E is the Electric Field, λ is the linear charge density, 𝜋 is the mathematical constant pi, ε0 is the permittivity of free space, and r is the distance from the line of charge.

What is the direction of the Electric Field due to line charge?

The direction of the Electric Field due to line charge is radial, meaning it points directly away from or towards the line of charge. The direction depends on the sign of the charge on the line and the position of the test charge relative to the line.

How does the Electric Field due to line charge change with distance?

The Electric Field due to line charge follows an inverse relationship with distance. This means that as the distance from the line of charge increases, the Electric Field decreases. This is because the Electric Field spreads out over a larger area as the distance increases.

How does the Electric Field due to line charge compare to other types of Electric Fields?

The Electric Field due to line charge is unique because it is a one-dimensional field, meaning it only varies in one direction. Other types of Electric Fields, such as those due to point charges or charged plates, are two-dimensional or three-dimensional fields. Additionally, the magnitude of the Electric Field due to line charge is dependent on the linear charge density, whereas the magnitude of other Electric Fields is dependent on the charge of the source and the distance from the source.

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