Question

A spherical shell has in inner radius R_i and an outer radius R_o. Within the shell, a total charge Q is uniformly distributed.

Calculate:

a) the charge density within the shell (if you cannot get this answer, you can proceed without it).

b) the electric field strength E(r) outside the shell (r > R_o).

c) the electric field strength inside the shell (r< R_i).

d) the electric field within the shell (R_i < r < R_o)

e) show that your solutions match both inner and outer boundaries

f) Draw a graph E versus r.

Answer 1

assume that :

R_i be the inner radius of the spherical shell.

R₀ be the outer radius of the spherical shell.

(a) the charge density within the shell is given as ::

= Q / V

(b) the electric field strength E(r) outside the shell (r > R_o) which is given as ::

according to gauss law,

. dA = Q_enclosed / ₀

E A = Q_enclosed / ₀

where, A = area of the sphere = 4r²

E (r) = (Q / 4₀ r²)

(c) the electric field strength inside the shell (r < R_i) which is given as ::

there is no charge, so the electric field is zero.     E (r) = 0

(d) the electric field within the shell (R_i < r < R_o) which is given as :

the total volume of the sphere is difference between two spheres of radii, R_i and R₀,

V = 4/3 (R₀³ - R_i³)       { eq. 1 }

and

= 3Q / 4(R₀³ - R_i³) { eq.2 }

Now, the gaussian surface encloses a volume which given as :

V_enclosed = (4/3) (r³- R_i³)                               { eq. 3 }

and enclosed charge is given as, Q_enclosed = V_enclosed    { eq. 4 }

inserting the values in eq.4,

Q_enclosed = 3Q / 4(R₀³ - R_i³) x (4/3) (r³- R_i³)

Q_enclosed = Q [(r³- R_i³) / (R₀³ - R_i³)]                                     { eq.4 }

using a gauss law,   . dA = Q_enclosed / ₀

E A = Q_enclosed / ₀

where, A = area of the sphere = 4r²

(r) = Q / 4₀ r² [(r³- R_i³) / (R₀³ - R_i³)]                                   (from eq. 4)

(e) the boundary conditions at both inner and outer radii which is given as :

when r = R_i which means that,

inserting the values in part-d eq.

(R_i) = Q / 4₀ R_i² [(R_i³- R_i³) / (R₀³ - R_i³)]

(R_i) = 0

it's value match with part-c.

when r = R₀ which means that,

again, using part-d eq :              (R₀) = Q / 4₀ R₀² [(R₀³- R_i³) / (R₀³ - R_i³)]

(R₀) = (Q / 4₀ R₀²)

these two conditions matching above solutions.

(f) plot a graph between E vs. r which can be shown in below ::