Question

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.

Answer 1

Answer: 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.