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/20 and points to the left. The electric field in the region to the right of B has magnitude 3σ/20 and points to the right. The electric field in the region between the two planes has magnitude σ/20 and points to the right. The surface charge density on planes A and B respectively is

The surface charge density on planes A and B respectively is

and

Explanation:

From the question we are told that

     The  electric field in region to the left of  A  is  

      The  direction of the electric field is  left

      The  electric field in the region to the right of  B  is  

      The  direction of the electric field is  right

      The  electric field in the region between the two planes is  

       The  direction of the electric field is  right

Let the surface charge density on planes A and B be represented as

 From the question we see that

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

=>      

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

=>      

=>      

From equation 1

=>    

So

=>  

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2