Question

Two planes of charge with no thickness, A and B, are parallel and vertical. The electric field in region I to the left of plane A has magnitude 3σ/2????0 and points to the left. The electric field in the region to the right of B has magnitude 3σ/2????0 and points to the right. The electric field in the region between the two planes has magnitude σ/2????0 and points to the right. The surface charge density on planes A and B respectively is

Answer 1

The surface charge density on planes A and B respectively is

`\sigma__(A)} } = 2\sigma`

and

`\sigma__(B)} = \sigma`

Step-by-step explanation:

From the question we are told that

The electric field in region to the left of A is
`E_i = (3 \sigma)/(2 \epsilon_o)`

The direction of the electric field is left

The electric field in the region to the right of B is
`E_f = (3 \sigma)/(2 \epsilon_o)`

The direction of the electric field is right

The electric field in the region between the two planes is
`E_m = (\sigma )/(2 \epsilon_o )`

The direction of the electric field is right

Let the surface charge density on planes A and B be represented as
`\sigma__(A)} \ \ and \ \ \sigma__(B)} \ \ \ respectively`

From the question we see that

`E_i = E_f`

Generally the electric to the right and to the left is due to the combined electric field generated by plane A and B so

`E_i = E_f = (3\sigma )/(2\epsilon) = (\sigma_A )/( 2 \epsilon_o ) + (\sigma_B )/( 2 \epsilon_o )`

=>
`\sigma__(A)} + \sigma__(B)} = 3 \sigma -- -(1)`

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

`(\sigma )/(2 \epsilon_o ) = (\sigma_A )/( 2 \epsilon_o ) - (\sigma_B )/( 2 \epsilon_o )`

=>
`\sigma__(A)} - \sigma__(B)} = \sigma`

=>
`\sigma__(A)} } = \sigma + \sigma__(B)`

From equation 1

`\sigma + \sigma__(B)}+ \sigma__(B)} = 3 \sigma`

=>
`\sigma__(B)} = \sigma`

So

`\sigma__(A)} } = \sigma + \sigma`

=>
`\sigma__(A)} } = 2\sigma`