Two 10-cm-diameter charged rings face each other, 18.0 cmcm apart. Both rings are charged to 30.0 nCnC . What is the electric field strength

E=7453.99 V/m

Explanation:

The electric field on the charged is given by

E= Kqx/(r^2 +x^2)^3/2

Where;

K= constant of Coulomb's law

q= magnitude of charge= 30.0×10^-9 C

r= radius of the rings= 5 cm or 0.05m

x= distance between the rings = 18cm = 0.18 m

Substituting values;

E= 9.0×10^9 × 30.0×10^-9 × 0.18 / [(0.05^2 + (0.18)^2]^3/2

E= 48.6/(2.5×10^-3 + 0.0324)^3/2

E= 48.6/(0.0025 + 0.0324)^3/2

E= 48.6/6.52×10^-3

E=7453.99 V/m

true is the correct answer

hope i