Net electric field in a circle

[jisbon]

Homework Statement

5 positive point charges are uniformly distributed and fixed around a circle of radius R. Determine the net electric field at the centre of the circle in terms of q and R.

Relevant Equations

-

In this case, I know there won't be any net efield in the x direction because it cancels out with each other.
The problem is dealing with the y axis. Am I supposed to presume an angle for each of them or what should I do instead?

Thanks

 

[Chestermiller]

Are you saying that you don't know enough Geometry to calculate the angles you need to know? What is 360/5?

 

[Orodruin]

I strongly suggest that you stop and think a bit before charging into compute angles. What would the field be if you replaced the 2q charge with a q charge?

 

[jisbon]

I strongly suggest that you stop and think a bit before charging into compute angles. What would the field be if you replaced the 2q charge with a q charge?

There won't be any net charge. Since it is 2q, I can assume the net efield is only a +q charge lying on the y axis?

 

[Orodruin]

There won't be any net charge.

I hope you mean net electric field, there certainly is a net charge and in any point other than the origin you cannot apply the symmetry argument.

Since it is 2q, I can assume the net efield is only a +q charge lying on the y axis?

You should not assume anythimg, you should argue for your conclusion. But yes, the electric field is additive. The total electric field will be that of five symmetrically positioned q charges and an additional q charge on the y axis.

 

[jisbon]

I hope you mean net electric field, there certainly is a net charge and in any point other than the origin you cannot apply the symmetry argument.You should not assume anythimg, you should argue for your conclusion. But yes, the electric field is additive. The total electric field will be that of five symmetrically positioned q charges and an additional q charge on the y axis.

Yep sorry I meant net efield. If there are 5 q charges positioned as shown, there won't be any net efield because the y and x-axis pretty much cancels out one another. So I can treat the 2q as two separate q charges, then conclude that the net efield is only due to a q charge on the y axis?