Two 10-cm-diameter charged rings face each other, 17.0 cm apart. Both rings are charged to + 40.0 nC . What is the electric field strength You may want to review (Pages 641 - 643) . For general problem-solving tips and strategies for this topic, you may want to view a Video Tutor Solution of Charge on a Van De Graaff.

each plate will experience an electric field strength of 12456.74N/C i.e 12.456 kN/C

Explanation:

Electric field strength E = F/Q

F is force of attraction or repulsion experienced by each charged ring due to the other charged ring

Q is the charge on each ring = 40nC

Also;

F = (QQ) /4¶£d^2

d is the distance between rings = 17cm

QQ = product of their charges

£ = permeability of free space

¶ is pi

From the above E becomes

E = Q/(4¶£d^2) as experienced by each ring.

E = (9*10^9 *40*10^-9) ÷ (17*10^-2)^2

NB 1/(4¶£) = 9*10^9

E = 360÷ 0.0289

E = 12457.74N/C

this is the electric field strength experienced by each ring.

the body's situation won't change the forces are equal opposit and on the same body

a.metals can become magnetic if they have many spinning electrons that create magnetic fields.d.only some metals are attracted by magnets; these metals are called ferromagnetic.

d.wrapping an insulated wire around a metal with ferromagnetic properties and applying an electric current.