Answer :
Answer:
The surface charge density on planes A and B respectively is
[tex]\sigma__{A}} } = 2\sigma[/tex]
and
[tex]\sigma__{B}} = \sigma[/tex]
Explanation:
From the question we are told that
The electric field in region to the left of A is [tex]E_i = \frac{3 \sigma}{2 \epsilon_o}[/tex]
The direction of the electric field is left
The electric field in the region to the right of B is [tex]E_f = \frac{3 \sigma}{2 \epsilon_o}[/tex]
The direction of the electric field is right
The electric field in the region between the two planes is [tex]E_m = \frac{\sigma }{2 \epsilon_o }[/tex]
The direction of the electric field is right
Let the surface charge density on planes A and B be represented as [tex]\sigma__{A}} \ \ and \ \ \sigma__{B}} \ \ \ respectively[/tex]
From the question we see that
[tex]E_i = E_f[/tex]
Generally the electric to the right and to the left is due to the combined electric field generated by plane A and B so
[tex]E_i = E_f = \frac{3\sigma }{2\epsilon} = \frac{\sigma_A }{ 2 \epsilon_o } + \frac{\sigma_B }{ 2 \epsilon_o }[/tex]
=> [tex]\sigma__{A}} + \sigma__{B}} = 3 \sigma -- -(1)[/tex]
Generally the electric field at the middle of the plane A and B is due to the diffencence in electric field generated by plane A and B
i.e
[tex]\frac{\sigma }{2 \epsilon_o } = \frac{\sigma_A }{ 2 \epsilon_o } - \frac{\sigma_B }{ 2 \epsilon_o }[/tex]
=> [tex]\sigma__{A}} - \sigma__{B}} = \sigma[/tex]
=> [tex]\sigma__{A}} } = \sigma + \sigma__{B}[/tex]
From equation 1
[tex]\sigma + \sigma__{B}}+ \sigma__{B}} = 3 \sigma[/tex]
=> [tex]\sigma__{B}} = \sigma[/tex]
So
[tex]\sigma__{A}} } = \sigma + \sigma[/tex]
=> [tex]\sigma__{A}} } = 2\sigma[/tex]