Electric Field due to line charge

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utkarshakash
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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) \\<br /> E_y = \dfrac{\lambda}{4 \pi \epsilon _0 d} |\cos 30 - \cos 60| \\<br /> <br /> \tan \alpha = \dfrac{E_y}{E_x} \\<br /> =\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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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.
 
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?
 
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:
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.
 
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 ?
 
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.
 
Fine . So a and b are integers . This is what I wanted to know .

Thanks a lot :)