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

Answer:6773.54 V/m

Explanation:

Given data

Distance between two rings(x)=25cm

Charge on each ring(Q)=$50\times 10^{-9}$ C

diameter of ring(d)=10 cm

Now Electric field strength at distance x from ring=$\frac{kxQ}{\left ( x^{2}+r^{2}\right )^{1.5}}$

Where K=coulomb's constant=$\frac{1}{4\pi \epsilon_0}$=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

$\left ( E\right )_{ring}$=$\frac{kxQ}{\left ( x^2+r^{2}\right )^{1.5}}$

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

$\left ( E\right )_{ring}$=6773.54 V/m