Use Gauss’ law to derive the expression for the electric field between two uniformly charge parallel sheets with surface charge densities σ and -σ respectively
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Answer:From triangle OAO', O
′
A=
R
2
−y
2
The charge on the length AB is q
ab
=2O
′
A×λ=2
R
2
−y
2
×λ
By Gauss's law, the electric flux ,ϕ=∮
E
.
dS
=
ϵ
0
q
enclosed
=
ϵ
0
q
ab
=
ϵ
0
2λ
R
2
−y
2
Explanation:
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