# Gaussian infinite sheet, electric field

## Homework Statement

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

## Homework Equations

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 ?

thanks

Last edited:

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Dick
Homework Helper
The strength of the electric field outside the infinite sheet is independent of the distance from the sheet. So the distance between the sheets doesn't matter when you are computing the E field. All you have to do is make sure you add the E fields in the correct directions. And, yes, for the potential difference the distance do matter. You integrate the E field over the separation between the sheets.

part b:

since electric potential difference V_ab = -[integral(E_net*dr)] from r_b to r_a where r_b = 6m, r_a = 0m and E_net = -1.41*10^12

V_ab = -1.41*10^12(6 - 0) = -8.46*10^12 volts

correct?

Dick