Question

In the figure there are three infinite long thin sheets having surface
charge density +2σ, -2σ and +σ respectively. Give the magnitude and
direction of electric field at a point to the left of sheet of charge density
+2σ and to the right of sheet of charge density +σ.

Answer 1

Answer:

The final answer is $\frac{\sigma}{2 \epsilon_{o}}(\hat{i})\end{aligned}$

Explanation:

Given:  There are three infinite long thin sheets having surface +2σ, -2σ, and +σ.

To Find : The magnitude and the direction of the electric field at a point to the left charge and charge density

+2σ and to the right charge density +σ.

Solution :

Gauss's law for electricity state that the electric flux Φ across any closed surface is proportional to the net electric charge q enclosed by the surface. That is, Φ = q/ε0, where ε0 is the electric permittivity and has a value of 8.854 × 10–12 square coulombs per newton per square meter.

$\begin{aligned}&\vec{E}_{A}=\frac{2 \sigma}{2 \epsilon_{o}}(-\hat{i})+\frac{2 \sigma}{2 \epsilon_{o}}(\hat{i})+\frac{\sigma}{2 \epsilon_{o}}(-\hat{i}) \\&=\frac{\sigma}{2 \epsilon_{o}}(-\hat{i}) \\&\vec{E}_{D}=\frac{\sigma}{2 \epsilon_{o}}(\hat{i})-\frac{2 \sigma}{2 \epsilon_{o}}(\hat{i})+\frac{2 \sigma}{2 \epsilon_{o}}(\hat{i}) \\&=\frac{\sigma}{2 \epsilon_{o}}(\hat{i})\end{aligned}$

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Answer 2

Concept

Surface charge density (σ) is the quantity of charge per unit area, measured in coulombs per square meter (C⋅m^2), at any point on a surface charge distribution on a two-dimensional surface.The formula of surface charge density is σ=q/A.

Given

We have given three infinite long thin sheets having surface charge densities +2σ, -2σ and +σ .

Find

We are asked to determine the electric field at the left of sheet of charge density +2σ and to the right of sheet of charge density +σ.

Solution

Electric field due to thin sheet of charge is given by E=σ/2ε0

Electric field at the left of sheet of charge density +2σ is = 2σ/2ε0 + (-2σ)/2ε0 + σ/2ε0

E = σ/2ε0

Electric field at right of sheet of charge density +σ is = 2σ/2ε0 + (-2σ)/2ε0 + σ/2ε0

E = σ/2ε0

Hence, Net electric field towards left is E = σ/2ε0 and

Net electric field towards right is E = σ/2ε0

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