Answer:
Part A:
[tex]E_{midpoint}=0[/tex]
Part B:
[tex]E_{center}=2711.7558 N/C[/tex]
Explanation:
Part A:
Formula of Electric Field Strength:
[tex]E=\frac{1}{4\pi\epsilon}\frac{xQ}{(x^2+R^2)^{3/2}}[/tex]
Where:
x is the distance from the ring
R is the radius of the ring
[tex]\epsilon[/tex] is constant permittivity of free space=8.854*10^-12 farads/meter
Q is the charge
For right Ring E at the midpoint can be calculated as:
x for right plate=25/2=12.5 cm=0.125 m
Radius=R=10/2=5 cm=0.05 m
[tex]E_{right}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0.125)*(20*10^{-19})}{((0.125)^2+(0.05)^2)^{3/2}}\\E_{right}=9208.1758 N/C[/tex]
For Left Ring E at the midpoint can be calculated as:
Since charge on both plates is +ve and same in magnitude, the electric field will be same for both plates.
[tex]E_{left}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0.125)*(20*10^{-19})}{((0.125)^2+(0.05)^2)^{3/2}}\\E_{left}=9208.1758 N/C[/tex]
Electric Field at midpoint:
Both rings have same magnitude but the direction of fields will be opposite as they have same charge on them.
[tex]E_{midpoint}=E_{left}-E_{right}\\E_{midpoint}=9208.1758-9208.1758\\E_{midpoint}=0[/tex]
Part B:
At center of left ring:
Due to left ring Electric field at center is zero because x=0.
[tex]E_{left}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0)*(20*10^{-19})}{((0)^2+(0.05)^2)^{3/2}}\\E_{left}=0 N/C[/tex]
Due to right ring Electric field at center of left ring:
Now: x=25 cm= o.25 m (To the center of left ring)
[tex]E_{right}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0.25)*(20*10^{-19})}{((0.25)^2+(0.05)^2)^{3/2}}\\E_{right}=2711.7558 N/C[/tex]
Electric Field Strength at center of left ring is same as that of right ring.
[tex]E_{center}=2711.7558 N/C[/tex]
