three thin charged sheets are placed parallel to each other as shown:
far left sheet, charge density = +10C/m^2
middle sheet, charge density = -25C/m^2
far right sheet, charge density = -10C/m^2
|<--2m (A) -->|<-----4m (B) ------>|
a) what is the electric field in regions A and B?
b) what is the difference in potential between the two outer sheets? neglect thickness of sheets
electric field at conductor surface = sigma/epsilon_o where sigma = charge density, epsilon_o = 8.85*10^-12 constant
electric flux = E (*) A where (*) is dot product, E is electric field, A is area of surface
charge enclose q_encl = sigma*A
The Attempt at a Solution
i've only tried part a:
what does it mean in the regions A and B, do i find the electric field of each sheet and then sum them up? how do distances between sheets factor in?
find the electric flux of the far left sheet, which is circle through the center of the sheet a la cookie cutter, and since there are two sides, there are two areas
electric flux = 2AE
sigma = +10C/m^2
q_encl = 10(2A)
let electric flux = q_encl/epsilon_o
so 2AE = sigma*A/epsilon_o ---> E = sigma/ 2*epsilon_o = 10/(2(8.85*10^-12)) = 5.65*10^11 N/C
that was the electric field for the far left sheet, i would do the same thing for the middle sheet but do i need to it to the far right sheet too for part a?
electric field middle sheet = -25/(2(8.85*10^-12)) = -1.41*10^12 N/C
electric field far right sheet = -10/(2(8.85*10^-12)) = -5.56*10^12 N/C
so the electric field for a and b = electric field( far left + middle + far right) = -1.41*10^12 N/C
how do the distances between the sheets come into play? ---> the 2m and 4m
also could you give me a quick run down on what i need to do for part b.
for part b since deltaV = -[integral(E*dr)] from r_a to r_b. do i let r_a = 0m and r_b = 6m ?