Question

Charge is placed on two conducting spheres that are very far apart and connected by a long thin wire. The radius of the smaller sphere is 5 cm and that of the larger sphere is 12 cm. The electric field at the surface of the larger sphere is 490 kV/m. Find the surface charge density on each sphere.

12 cm sphere
? µC/m^2

5 cm sphere
? µC/m^2

Answer 1

Concepts and reason

The required concepts to solve this problem are electric field and surface charge density.

First, from the equations of electric field by the smaller and larger spheres, find the ratio of the electric field by the smaller sphere to the electric field by the larger sphere. The by equating the potentials of both the spheres, find the ratio of charges of the smaller sphere and the larger sphere and finally, find the electric field by the smaller sphere. Using the electric field of the smaller and larger sphere, find the surface charge density of both.

Fundamentals

The surface charge density is the electric charge per unit area. Surface charge density can be represented in equation form as

$\sigma = \frac{Q}{A}$

Here, $Q$is the charge and$A$is the area.

Surface charge density of a sphere is,

$\sigma = {\varepsilon _o}E$

Here, $E$is the electric field at the surface.

The electric field of a conducting sphere is,

$E = \frac{1}{{4\pi {\varepsilon _o}}}\frac{Q}{{{R^2}}}$

Here, $Q$is the charge at the surface of the sphere and$R$ is the radius of the sphere.

The potential of a conducting sphere is,

$V = \frac{1}{{4\pi {\varepsilon _o}}}\frac{Q}{R}$

Here, $Q$is the charge at the surface of the sphere and$R$ is the radius of the sphere.

The equation for the surface charge density of the larger sphere is,

${\sigma _L} = {\varepsilon _o}{E_L}$

Substitute $8.854 \times {10^{ - 12}}\,{\rm{F/m}}$for ${\varepsilon _o}$ and $490\,{\rm{kV/m}}$for ${E_L}$to find${\sigma _L}$.

$\begin{array}{c}\\{\sigma _L} = \left( {8.854 \times {{10}^{ - 12}}\,{\rm{F/m}}} \right)\left( {\left( {490\,{\rm{kV/m}}} \right)\left( {\frac{{{{10}^3}\,{\rm{V}}}}{{1\,{\rm{kV}}}}} \right)} \right)\\\\ = 4.34 \times {10^{ - 6}}\,{\rm{C/}}{{\rm{m}}^2}\\\\ = 4.34\,{\rm{\mu C/}}{{\rm{m}}^{\rm{2}}}\\\end{array}$

The equation for electric field of the larger sphere is,

${E_L} = \frac{1}{{4\pi {\varepsilon _o}}}\frac{{{Q_L}}}{{R_L^2}}$

The equation for electric field of the smaller sphere is,

${E_S} = \frac{1}{{4\pi {\varepsilon _o}}}\frac{{{Q_S}}}{{R_S^2}}$

The ratio of electric field of smaller and larger spheres is,

$\frac{{{E_S}}}{{{E_L}}} = \left( {\frac{{{Q_S}}}{{{Q_L}}}} \right)\left( {\frac{{R_L^2}}{{R_S^2}}} \right)$ …… (1)

The two spheres are connected with conducting wire. Thus, both the spheres have the same electric potential.

$\begin{array}{l}\\{V_S} = {V_L}\\\\\frac{1}{{4\pi {\varepsilon _o}}}\frac{{{Q_S}}}{{{R_S}}} = \frac{1}{{4\pi {\varepsilon _o}}}\frac{{{Q_L}}}{{{R_L}}}\\\end{array}$

$\frac{{{Q_S}}}{{{Q_L}}} = \frac{{{R_S}}}{{{R_L}}}$ …… (2)

Substituting equation$\left( 2 \right)$in equation$\left( 1 \right)$, the electric field ratio is,

$\begin{array}{c}\\\frac{{{E_S}}}{{{E_L}}} = \left( {\frac{{{R_S}}}{{{R_L}}}} \right)\left( {\frac{{R_L^2}}{{R_S^2}}} \right)\\\\ = \frac{{{R_L}}}{{{R_S}}}\\\end{array}$

So, the equation for electric field of the smaller sphere is,

${E_S} = \left( {\frac{{{R_L}}}{{{R_S}}}} \right){E_L}$

Substitute$12\,{\rm{cm}}$for${R_L}$, $5\,{\rm{cm}}$for${R_S}$and $490\,{\rm{kV/m}}$for ${E_L}$to find${E_S}$.

$\begin{array}{c}\\{E_S} = \left( {\frac{{12\,{\rm{cm}}}}{{5\,{\rm{cm}}}}} \right)\left( {490\,{\rm{kV/m}}} \right)\\\\ = 1176\,{\rm{kV/m}}\\\end{array}$

The equation for the surface charge density of the smaller sphere is,

${\sigma _S} = {\varepsilon _o}{E_S}$

Substitute $8.854 \times {10^{ - 12}}\,{\rm{F/m}}$for ${\varepsilon _o}$ and $1176\,{\rm{kV/m}}$for ${E_S}$to find${\sigma _S}$.

$\begin{array}{c}\\{\sigma _L} = \left( {8.854 \times {{10}^{ - 12}}\,{\rm{F/m}}} \right)\left( {\left( {1176\,{\rm{kV/m}}} \right)\left( {\frac{{{{10}^3}\,{\rm{V}}}}{{1\,{\rm{kV}}}}} \right)} \right)\\\\ = 10.4 \times {10^{ - 6}}\,{\rm{C/}}{{\rm{m}}^2}\\\\ = 10.4\,{\rm{\mu C/}}{{\rm{m}}^{\rm{2}}}\\\end{array}$

Ans:

The surface charge density of the large sphere is$4.34\,{\rm{\mu C/}}{{\rm{m}}^{\rm{2}}}$

The surface charge density of the large sphere is$10.4\,{\rm{\mu C/}}{{\rm{m}}^{\rm{2}}}$.