Electric Fields from Linear Charges

[Mark Zhu]

Homework Statement

Am amount of charge Q is uniformly spread over a semi-circle of radius R whose center is located a distance A from the origin. What point charge would have to be placed at the origin so that the E field at the center of the circle is 0? (The open end of the semi-circle is facing away from the origin, and is symmetric about the x axis).

Homework Equations

E = (k*q)/(r^2)
λ = Q/(pi*R)
λ = dq/ds
ds = R*dθ

The Attempt at a Solution

First we break up the entire line of charge into tiny bits of charge 'dq'. We look at a single 'dq' and consider the tiny charge's contribution to the electric field at point 'A' (the center of the semi-circle) in the x and y directions. We integrate 'dEx' and 'dEy' both with respect to 'θ' over the entire semi-circle of charge. Of course, we have to first do a series of substitutions for some of the variables using the above "relevant equations", before we can actually do the two integrations. Then we get 'Ex' and 'Ey', and all is well.

My question is since in this problem 'Ey' turns out to be zero due to symmetry, and so 'Ex' equals 'E', do we even need to break up 'dE' into its x and y components? If not, then in the integral of 'dEx' we could leave out the annoying 'sin θ' thus simplifying the calculation.

 

[kuruman]

My question is since in this problem 'Ey' turns out to be zero due to symmetry, and so 'Ex' equals 'E', do we even need to break up 'dE' into its x and y components? If not, then in the integral of 'dEx' we could leave out the annoying 'sin θ' thus simplifying the calculation.

You can invoke symmetry and that should be sufficient. However, you can also do the integral for E_y formally and verify that it is zero. Either way is correct.

 

[SammyS]

Homework Statement

Am amount of charge Q is uniformly spread over a semi-circle of radius R whose center is located a distance A from the origin. What point charge would have to be placed at the origin so that the E field at the center of the circle is 0? (The open end of the semi-circle is facing away from the origin, and is symmetric about the x axis).

The Attempt at a Solution

...

My question is since in this problem 'Ey' turns out to be zero due to symmetry, and so 'Ex' equals 'E', do we even need to break up 'dE' into its x and y components? If not, then in the integral of 'dEx' we could leave out the annoying 'sin θ' thus simplifying the calculation.

You must include the "annoying" $\sin\theta$ to get $E_x ~ .$

 

[Riyad Rahman]

tHANKS FOR THE HELPFUL POST!

 

[Delta2]

You must include $\sin\theta$ to get the correct expression for $E_x$. The correct contribution to $E_x$ from each dq includes the $\sin\theta$ term, otherwise you would integrating the wrong contribution from each dq and thus you would getting the wrong result for $E_x$.

Because the contributions to $E_y$ sum up to zero due to symmetry, this doesn't give you the right "to transfer" the $E_y$ contributions to $E_x$ contributions and thus omit the $\sin\theta$ term.

 

[Mark Zhu]

Thank you for all your great help guys, I understand it now.

 

[Np14]

oh

 

[PhysicsKT]

Tip: Doing this in polar coordinates should help. After you've cart bashed it, try by polars too.