Question

Two 10-cm-diameter charged rings face each other, 25.0 cm apart. Both rings are charged too + 50.0 nC . What is the electric field strength?

Answer 1

The electric field strength between the two charged rings is approximately 1.49 x 10^7 N/C.

Step-by-step explanation:

The electric field strength between the two charged rings can be calculated using Coulomb's law. The formula for the electric field strength, E, due to a charged ring is given by:

E = (kQ) / (2πεr)

Where k is the electrostatic constant (8.99 x 10^9 N·m^2/C^2), Q is the charge on the ring (50.0 nC), ε is the permittivity of free space (8.85 x 10^-12 C^2/N·m^2), and r is the distance between the rings (25.0 cm).

Substituting the given values into the formula, we have:

E = [(8.99 x 10^9 N·m^2/C^2) * (50.0 x 10^-9 C)] / [2π(8.85 x 10^-12 C^2/N·m^2)(0.25 m)]

E ≈ 1.49 x 10^7 N/C

Therefore, the electric field strength between the two charged rings is approximately 1.49 x 10^7 N/C.

Answer 2

Answer:6773.54 V/m

Step-by-step explanation:

Given data

Distance between two rings(x)=25cm

Charge on each ring(Q)=
`50* 10^(-9)` C

diameter of ring(d)=10 cm

Now Electric field strength at distance x from ring=
`(kxQ)/(\left ( x^(2)+r^(2)\right )^(1.5))`

Where K=coulomb's constant=
`(1)/(4\pi \epsilon_0)`=8.98\times 10^{9]

Now electric field strength at midpoint between two rings is zero

because one is pointing towards positive x direction and other in negative z direction.

but electric field strength at center of ring due to other ring is given by

`\left ( E\right )_(ring)`=
`(kxQ)/(\left ( x^2+r^(2)\right )^(1.5))`

`\left ( E\right )_(ring)`=
`\frac{8.98* 10^(9)* 0.25* 50* 10{-9}}\left ( 0.25^2+0.05^(2)\right )}^(1.5)}`

`\left ( E\right )_(ring)`=6773.54 V/m