Answer:
Answer is explained in the explanation section below.
Explanation:
Solution:
We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.
So,
a) 0 < r < r1 :
We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.
Hence, E = 0 for r < r1
b) r1 < r < r2:
Electric field =?
Let, us consider the Gaussian Surface,
E x 4 =
So,
Rearranging the above equation to get Electric field, we will get:
E =
Multiply and divide by
E = x
Rearranging the above equation, we will get Electric Field for r1 < r < r2:
E= (σ1 x ) /( x )
c) r > r2 :
Electric Field = ?
E x 4 =
Rearranging the above equation for E:
E =
E = +
As we know from above, that:
= (σ1 x ) /( x )
Then, Similarly,
= (σ2 x ) /( x )
So,
E = +
Replacing the above equations to get E:
E = (σ1 x ) /( x ) + (σ2 x ) /( x )
Now, for
d) Under what conditions, E = 0, for r > r2?
For r > r2, E =0 if
σ1 x = - σ2 x