# A solid plastic sphere of radius R1 = 0.300 m is contained inside a thin spherical shell of...

## Question:

A solid plastic sphere of radius R1 = 0.300 m is contained inside a thin spherical shell of radius R2 = 0.600 m and is concentric with it. In between and concentric is a thick spherical metal conducting shell of inner radius 0.400 m and outer radius 0.500 m with no net charge. The innermost shell carries total charge Q1 distributed uniformly throughout its volume. The outermost shell carries total charge Q2 distributed uniformly on its surface. At r = 0.550 m, the electric field is 53,500 N/C radially outward.

1. At r = 0.650 m, the electric field is 76,609 N/C radially outward. What are Q1 and Q2?

2. Find the electric field at the given value of r, where r is the distance measured from the centers.

a. r = 0.150 m

b. r = 0.250 m

c. r = 0.350 m

d. r = 0.450 m

3. Find the surface charge densities on the inner and outer surfaces of the metal conducting shell.

## Electric Field

Electric field due to a point charge or any charge distribution can be calculated using the Gauss law in electrostatics. Gauss law in electrostatics says that electric flux through a closed surface is equal to the net charge *Q* enclosed by the surface divided by the permittivity of free space {eq}\epsilon_0 {/eq}. Electric flux {eq}\phi_e = \dfrac { Q } { \epsilon_0 } {/eq}. Electric flux is the dot product of electric field *E* and area *A* through which the electric field is passing through. Electric field is a vector quantity having both magnitude and direction As a result electric field obeys the principle of superposition.

- Electric field at a point can be calculated using Gauss law and superposition principle.
- Electric field can induce charge on the surface of conductors and insulators.
- Electrostatic field does not exist inside a conductor.

## Answer and Explanation:

**Given data**

- Radius of the innermost solid plastic sphere {eq}R_1 = 0.300 \ m {/eq}

- The radius of the concentric spherical shell surrounding the solid sphere {eq}R_2 = 0.600 \ m {/eq}

- A thick conducting concentric spherical shell is there in between the solid sphere and outer shell with no net charge deposited on it.
- Inner radius of the conducting shell {eq}r_1 = 0.400 \ m {/eq}

- Outer radius of the conducting shell {eq}r_2 = 0.500 \ m {/eq}

- The inner most solid sphere has a net charge {eq}Q_1 {/eq} distributed uniformly through out the volume and the outermost shell has a net charge {eq}Q_2 {/eq} uniformly distributed uniformly on the outer surface.

- Electric field at a radial distance {eq}r_a = 0.550 \ m {/eq} is {eq}E_a = 53,500 \ N/C {/eq} directed radially outwards.

- Electric field at radial distance {eq}r_b = 0.650 \ m {/eq} is {eq}E_b = 76,609 \ N/C {/eq}

- Coulomb's constant {eq}k = 8.99 \times 10^9 \ N . m^2 /C^2 {/eq}

**Part 1 )**

Electric field {eq}E_a {/eq} is due to the enclosed charge {eq}Q_1 {/eq} uniformly distributed in the volume of the solid sphere. This charge induces charge on the conducting shell and its field is available outside the conducting shell

Therefore the electric field {eq}E_a = \dfrac { k Q_1 } { r_a^2 } {/eq}

Therefore the charge {eq}Q_1 = \dfrac { E_a \times r_a^2} { k } \\ Q_1 = \dfrac { 53500 \times 0.550^2 } { 8.99 \times 10^9 } \\ Q_1 = +1.80 \times 10^{-6} \ C {/eq}

Electric field {eq}E_b {/eq} is due to both the charges.

Then electric field {eq}E_b = \dfrac { k ( Q_1 + Q_2 ) } { r_b^2 } {/eq}

Therefore the sum of the charges {eq}Q_1 + Q_2 = \dfrac { E_b \times r_b^2 } { k } \\ Q_1 + Q_2 = \dfrac { 76609 \times 0.650^2 } { 8.99 \times 10^9 } \\ Q_1 + Q_2 = 3.60 \times 10^{ -6} \ C {/eq}

Therefore the value charge {eq}Q_2 = 3.600 \times 10^{-6} - 1.80 \times 10^{-6} \\ Q_2 = +1.80 \times 10^{-6} \ C {/eq}

**Part 2)**

Volume charge density in the solid sphere {eq}\rho = \dfrac { Q_1 } { V } \\ \rho = \dfrac { 3 Q_1 } { 4 \pi R_1 ^3 } {/eq}

**a)**

Electric field at radial distance {eq}r = 0.150 \ m {/eq}

Electric field {eq}E_1 = \dfrac { k \rho V_1 } { r^2 } \\ E_1 = \dfrac { k Q_1 V_1 } { V r^2 } \\ E_1 = \dfrac { k Q_1 r^3 } { R_1^3 r^2 } \\ E_1 = \dfrac { k Q_1 r } { R_1^3 } \\ E_1 = \dfrac { 8.99 \times 10^9 \times 1.8 \times 10^{-6} \times 0.150 } { 0.300^3 } \\ E_1 = 89900 \ N/C {/eq}

**b)**

Electric field at {eq}r = 0.250 \ m {/eq}

Electric field at the given point {eq}E_2 = \dfrac { k Q_1 r } { R_1^3 } \\ E_2 = \dfrac { 8.99 \times 10^9 \times 0.250 } { 0.300^3 } \\ E_2 = 149833 \ N/C {/eq}

**c)**

Electric field at radial distance {eq}r = 0.350 \ m {/eq}

The Gaussian surface of radius *r* encloses the whole solid sphere.

Therefore the electric field at the given point {eq}E_3 = \dfrac { k Q_1 } { r^2 } \\ E_3 = \dfrac { 8.99 \times 10^9 \times 1.80 \times 10^{-6} } { 0.350^2 } \\ E_3 = 132098 \ N/C {/eq}

**d)**

Electric field at radial distance {eq}r = 0.450 \ m {/eq}.

The Gaussian surface of this radius passes through the thick conducting shell and hence no electric field at this point.

**Part 3 )**

Charge induced on the inner surface of the thick conducting shell {eq}q_{d1} = - Q_1 {/eq}

Therefore surface charge density on the inner surface of the conducting shell {eq}\sigma_1 = \dfrac { - Q_1 } { 4 \pi r_a^2 } \\ \sigma_1 = \dfrac { - 1.80 \times 10^{-6} } { 4 \pi 0.400^2 } \\ \sigma_1 = - 8.95 \times 10^{-7} \ C/ m^2 {/eq}

Charge induced on the outer surface of the conducting shell {eq}q_{d2} = Q_1 {/eq}

Surface charge density on the outer surface of the conducting shell {eq}\sigma_2 = \dfrac { Q_1 } { 4 \pi r_b^2 } \\ \sigma_2 = \dfrac { 1.80 \times 10^{-6} } { 4 \pi 0.500^2 } \\ \sigma_2 = 5.73 \times 10^{-7} \ C/ m^2 {/eq}