Question

Three charged metal disks are arranged as shown (cutaway view). The disks are held apart by insulating supports not shown in the diagram. Each disk has an area of 2.5 m2 (this is the area of one flat surface of the disk). Use the value 8.85e-12 C2/(N·m2) for e0.The charge Q1 = 3e-8 coulombs, and the charge Q2 = 2e-7 coulombs.

The figure of the disks can be viewed at this website: http://www.webassign.net/mi3/17-070-3plates_D_res.jpg.

What is the potential difference between VC-VB and VD-VF?

Answer 1

their are no correct answers, I am not sure how to answer that but I would like to learn how?

Answer 2

Magnitude of electric field on a metallic charge conductor = sigma/2Epsilon_0
where sigma is the charge density of the plate
sigma 1 = 3*10^-8/2.5
sigma 2 = 2 *10^-7 /2.5

Electric field of plate 1 = 3*10^-8/(2*2.5*8.85e-12)
Electric field of plate 2 left side = sigma/Epsilon_0
= -3*10^-8/(2.5*8.85e-12)
Electric field of plate 2 right side = -2 *10^-7 /(2.5*8.85e-12)
Electric field of plate 3 = 2 *10^-7/(2*2.5*8.85e-12)

Electric field in region left = 1.5*3*10^-8/(2.5*8.85e-12) directed right
Electric field in region right = 1.5*2 *10^-7 /(2.5*8.85e-12)

Vc-Vb = 1.5*3*.004*10^-8/(2.5*8.85e-12) = 8.135593 volts
Vd- Vf = .0017*1.5*2 *10^-7 /(2.5*8.85e-12) = 23.050847 volts