Question

Two 10-cm-diameter charged rings face each other, 25.0cm apart. Both rings are charged to +20.0nC. What is the electric field strength at:

a) the midpoint between the two rings?
b) the center of the left ring?

Answer 1

Part A:

`E_(midpoint)=0`

Part B:

`E_(center)=2711.7558 N/C`

Step-by-step explanation:

Part A:

Formula of Electric Field Strength:

`E=(1)/(4\pi\epsilon)(xQ)/((x^2+R^2)^(3/2))`

Where:

x is the distance from the ring

R is the radius of the ring

`\epsilon` 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

`E_(right)=(1)/(4\pi8.854*10^(-12))((0.125)*(20*10^(-19)))/(((0.125)^2+(0.05)^2)^(3/2))\\E_(right)=9208.1758 N/C`

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.

`E_(left)=(1)/(4\pi8.854*10^(-12))((0.125)*(20*10^(-19)))/(((0.125)^2+(0.05)^2)^(3/2))\\E_(left)=9208.1758 N/C`

Electric Field at midpoint:

Both rings have same magnitude but the direction of fields will be opposite as they have same charge on them.

`E_(midpoint)=E_(left)-E_(right)\\E_(midpoint)=9208.1758-9208.1758\\E_(midpoint)=0`

Part B:

At center of left ring:

Due to left ring Electric field at center is zero because x=0.

`E_(left)=(1)/(4\pi8.854*10^(-12))((0)*(20*10^(-19)))/(((0)^2+(0.05)^2)^(3/2))\\E_(left)=0 N/C`

Due to right ring Electric field at center of left ring:

Now: x=25 cm= o.25 m (To the center of left ring)

`E_(right)=(1)/(4\pi8.854*10^(-12))((0.25)*(20*10^(-19)))/(((0.25)^2+(0.05)^2)^(3/2))\\E_(right)=2711.7558 N/C`

Electric Field Strength at center of left ring is same as that of right ring.

`E_(center)=2711.7558 N/C`