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A spherical shell of radius Ra carries a total charge qa uniformly distributed on its surface. A larger spherical shell of radius Rb is concentric with the first and carries a charge qb uniformly distributed on its surface. (a) Use Gauss’ law to find E~ in all regions (do not forget the inside of the smaller shell!). (b) If qa=6nC, what should qb be for the electric field to be zero for r > Rb? (c) Sketch the electric-field lines for the situation in part (b).

Respuesta :

Answer:

Explanation:

Part a)

For a position of point inside the inner shell we can use Gauss law as

[tex]\int E. dA = \frac{q}{\epsilon_0}[/tex]

now here we know that enclosed charge in the inner shell is ZERO

so we have

[tex]\int E. dA = 0[/tex]

[tex]E = 0[/tex]

Now for the position between two shells

[tex]r_a< r< r_b[/tex]

again by Gauss law

[tex]\int E. dA = \frac{q}{\epsilon_0}[/tex]

now here we know that enclosed charge between two shells is given as

[tex]q = q_a[/tex]

so we have

[tex]\int E. dA = \frac{q_a}{\epsilon_0}[/tex]

[tex]E = \frac{q_a}{4\pi \epsilon_0 r^2}[/tex]

Now for position outside the shell we will have

[tex]r > r_b[/tex]

again by Gauss law

[tex]\int E. dA = \frac{q}{\epsilon_0}[/tex]

now here we know that enclosed charge given as

[tex]q = q_a + q_b[/tex]

so we have

[tex]\int E. dA = \frac{q_a + q_b}{\epsilon_0}[/tex]

[tex]E = \frac{q_a + q_b}{4\pi \epsilon_0 r^2}[/tex]

Part b)

If outside the shell net electric field is zero

then we can say

[tex]q_a + q_b = 0[/tex]

[tex]q_a = 6 nC[/tex]

[tex]q_b = - 6nC[/tex]

Part c)

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