Question

b) In the figure there are three infinite long thin sheets having surface
charge density +20, -20 and to respectively. Give the magnitude and
direction of electric field at a point to the left of sheet of charge density
+20 and to the right of sheet of charge density +o.
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bis 15 nap imo o nolastnini sido
10 ar 20 U-20 reoler
nu sinni var
A
in swanie (6
o ene woda
opsiberi wa
B Бас
D
To
10 Susi
Honia
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Answer 1

Given:

no of sheet N = 3

charge density σ = 20 c$m^2$

To find:

electric field E =?

Explanation:

• Gauss law is the total flux linked with a closed surface is 1/ε0 times the charge enclosed by the closed surface.

• It mathamatically expressed as

E ds = $\displaystyle \frac{1}{\epsilon_0}$ qN       ...(1)

where

E = electric field

ds = surface area

$\epsilon_0$ =  the electric constant

q = charge

• But charge density is the charge flowing through unit area.

∴ $\displaystyle \sigma = \frac{qN}{ds}$

• Put this in equation (1) we get,

$\displaystyle E = \frac{\sigma}{\epsilon_0}$

   =   $\displaystyle \frac{20\times 3}{8.85\times10^{-12}}$

E = 7.5 TV/m