Question

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

Answer 1

Complete 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:

a) the electric field strength at the midpoint between the two rings is 0

b) the electric field strength at the center of the left ring is 2712.44 N/C

Step-by-step explanation:

Given;

distance between the two rings, d = 25 cm = 0.25 m

diameter of each ring, d = 10 cm = 0.1 m

radius of each ring, r =
`(0.1)/(2) = 0.05 \ m`

the charge on each ring, q = 20 nC

Electric field strength for a ring with radius r and distance x from the center of the ring is given as;

`E = (kxQ)/((x^2 +r^2)^(3/2))`

The electric field strength at the midpoint;

the distance from the left ring to the mid point , x = 0.25 m / 2 = 0.125 m

`E = (kxQ)/((x^2 +r^2)^(3/2)) \\\\E = (8.99*10^(9)*0.125*20*10^(-9))/((0.125^2 + 0.05^2)^(3/2)) \\\\E = 9210.5 \ N/C`

`E_(left) = 9210.5 \ N/C`

The electric field strength due to right ring is equal in magnitude to left ring but opposite in direction;

`E_(right) = -9210.5 \ N/C`

The electric field strength at the midpoint;

`E_(mid) = E_(left) + E_(right)\\\\E_(mid) = 9210.5 \ N/C - 9210.5 \ N/C\\\\E_(mid) = 0`

(b)

The distance from the right ring to center of the left ring, x = 0.25 m.

`E = (KxQ)/((x^2 +r^2)^(3/2)) \\\\E = (8.99*10^(9) *0.25*20*10^(-9))/((0.25^2 + 0.05^2)^(3/2)) \\\\E = 2712.44 \ N/C`