Question

The three parallel planes of charge shown in the figure (Figure 1) have surface charge densities −12 η, η, and −12 η. Find the magnitude in each of the 4 reigons.

Answer 1

Electric field =
`E = -(\pi )/(2\mu )`

Step-by-step explanation:

Thinking process:

Because r r E = 0 in the metal there will be an induced charge polarization. The face of the conductor adjacent to the plane of charge is negatively charged. This makes the other face of the conductor positively charged. We thus have three infinite planes of charge. These are P (top conducting face), P′ (bottom conducting face), and P″(plane of charge).

Let η1, η2, and η3 be the surface charge densities of the three surfaces with η2 a negative number. The electric field due to a plane of charge with surface charge density η is E = η ε2 0 . Because the electric field inside a conductor is zero (region 2)

We have made the substitution η3 = η. Also note that the field inside the conductor is downward from planes P and

P′ and upward from P″. Because η1 + η2 = 0 C/m2

because the conductor is neutral, η2 = −η1.

The above equation becomes :

`E = -(2\pi )/(\mu )`