The figure shows a spherical shell with uniform volume charg

The figure shows a spherical shell with uniform volume charge density row=90 C/m^3, inner radius a= 15 cm, and outer radius b=30. Find the magnitude of the electric field at the following radial distances: r=0 r=10cm r=15cm. r=25cm r=30cm. r=45cm

Solution

To calculate the magnitude of electric field inside the sphere, we are given with the density () and radius of the two spheres. We are asked to calculate the electric field at various radial distances for which we need to calculate the electric field on a given distance due b radius sphere:

we have formula to compute the electric field as:-

E.dA (over closed surface) = Q(inside)/(0)           ...(1)

where E is the electric field and dA is the small area , Q is the charge inside the spherical shell, (0) is the constant.

To calculate the Q(inside) consider a small surface inside the given sphere with radius r consider a smaller area of this sphere as da, dr and dV(volume) respectively.
   dQ = ( (0) (r^2/b^2) * (dV)
         = ( (0) (r^2/b^2) * (4*pi*r^2)dr
         = ((0)*4*pi*(r^4/b^2) dr
         = (0)*4*pi / b^2    (r^4)dr   (limits of integration lower limit 0, and upper limit value of r. Given in the Question)

which gives us:-

dQ =   (4*pi* * r^5)/5b^2       ...(2)

now putting this in eq 1
we get, final answer as:

E = *r^3 / (5*b^2*(0) )       ...(3)

For case 1:

r = 0 , b = 30cm = .3 m, = 30 nC/m^3 = 9e-8 C/m^3
if we put values in eq 2 and 3, we get E=0

for case 2:
r = 10cm = 0.1m , b = 30cm = 0.3 m, = 30 nC/m^3 = 9e-8 C/m^3
if we put values eq 3,

E = *r^3 / (5*b^2*(0) )
E = 9e-8 *0.1^3 / (5*0.3^2*9*10^9 )
we get E=24.464536456

for case 3:
r = 15cm = 0.15m , b = 30cm = 0.3 m, = 30 nC/m^3 = 9e-8 C/m^3
if we put values eq 3,

E = *r^3 / (5*b^2*(0) )
E = 9e-8 *0.15^3 / (5*0.3^2*9*10^9 )
we get E=24.464536456

for case 4:
r = 25cm = 0.25m , b = 30cm = 0.3 m, = 30 nC/m^3 = 9e-8 C/m^3
if we put values eq 3,

E = *r^3 / (5*b^2*(0) )
E = 9e-8 *0.25^3 / (5*0.3^2*9*10^9 )
we get E=24.464536456

for case 5:
r = 25cm = 0.25m , b = 30cm = 0.3 m, = 30 nC/m^3 = 9e-8 C/m^3
if we put values eq 3,

E = *r^3 / (5*b^2*(0) )
E = 9e-8 *0.25^3 / (5*0.3^2*9*10^9 )
we get E=24.464536456

Now we need to calculate the Electric field outside the sphere:
whcih can be calculated as:

E= *b^3 / (5*r^2*(0) )         ...(4)
for case 6:
r = 45cm = 0.45m , b = 30cm = 0.3 m, = 30 nC/m^3 = 9e-8 C/m^3
if we put values eq 4,

E = 9e-8 *0.30^3 / (5*0.45^2*9*10^9 )
we get E=24.464536456