Answer:
[tex]Electric\ field = \dfrac{4KQ }{\pi R^2}[/tex]
Explanation:
Given that
Top rod have +4Q charge
Bottom tube have -2Q charge
We know that electric filed due to half ring on the center given as
[tex]E=\dfrac{2K\lambda }{R}[/tex]
Where
λ is the charge per unit length
K is the constant
R is the radius of ring
For top rod
λ=+4Q/πR
So
[tex]E=\dfrac{2K\times 4Q }{\pi R^2}[/tex]
[tex]E=\dfrac{8KQ }{\pi R^2}[/tex]
For bottom ring
[tex]E'=-\dfrac{2K\times 2Q }{\pi R^2}[/tex]
[tex]E'=-\dfrac{4KQ }{\pi R^2}[/tex]
So the total resultant electric field at the center= E-E'
[tex]Electric\ field = \dfrac{8KQ }{\pi R^2}-\dfrac{4KQ }{\pi R^2}[/tex]
[tex]Electric\ field = \dfrac{4KQ }{\pi R^2}[/tex]
