Question

An isolated charged conducting sphere of radius 10.0 cm creates an electric field of 4.90 10⁴ N/C at a distance 23.0 cm from its center.

(a) What is its surface charge density?
µC/m²

(b) What is its capacitance?
pF

Answer 1

Concepts and reason

The concepts used to solve this problem are electric field due to conducting sphere, the surface area of the sphere, surface charge density and capacitance of the isolated spherical capacitor.

Use the expression of the electric field due to conducting sphere to calculate the charge on the sphere.

Use the expression for the surface area of the sphere to calculate the surface area of the sphere.

Use the expression of surface charge density to calculate the surface charge density of the sphere.

Use the expression for the capacitance of the isolated spherical capacitor to calculate the capacitance.

Fundamentals

Write the expression for the electric field due to conducting sphere.

$E = \frac{{kq}}{{{r^2}}}$

Here, $k$ is the electrostatic constant, $q$ is the charge on the conducting sphere and $r$ is the distance at which the electric field is evaluated from the center of the sphere.

Write the expression for the surface area of a sphere.

$A = 4\pi {R^2}$

Here, $R$ is the radius of conducting sphere.

Write the expression for surface charge density.

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

Here, $q$ is the charge on the conducting sphere and $A$ is the surface area of the sphere.

Write the expression for the capacitance of the isolated spherical capacitor.

$C = 4\pi {\varepsilon _0}R$

Here, ${\varepsilon _0}$is the absolute permittivity and $R$ is the radius of the sphere.

(a)

Calculate the charge on the sphere.

Write the expression for the electric field due to conducting sphere.

$E = \frac{{kq}}{{{r^2}}}$

Here $k$ is the electrostatic constant, $q$ is the charge on the conducting sphere and $r$ is the distance at which the electric field is evaluated from the center of the sphere.

Rearrange for the charge.

$q = \frac{{E{r^2}}}{k}$

Substitute ${\rm{4}}{\rm{.9 \times 1}}{{\rm{0}}^{\rm{4}}}{\rm{ N/C}}$ for $E$, ${\rm{9 \times 1}}{{\rm{0}}^{\rm{9}}}{\kern 1pt} {\rm{ N}} \cdot {\kern 1pt} {{\rm{m}}^{{\rm{2}}{\kern 1pt} }}{\rm{/}}{{\rm{C}}^2}$ for $k$ and $23{\kern 1pt} {\rm{ cm}}$for $r$.

$\begin{array}{c}\\q = \frac{{\left( {4.9 \times {{10}^4}{\rm{ N/C}}} \right){{\left( {23{\rm{ cm}}} \right)}^2}}}{{\left( {9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right)}}\\\\ = \frac{{\left( {4.9 \times {{10}^4}{\rm{ N/C}}} \right){{\left( {\left( {23{\rm{ cm}}} \right)\left( {\frac{{1{0^{ - 2}}{\rm{ m}}}}{{1{\rm{ cm}}}}} \right)} \right)}^2}}}{{\left( {9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right)}}\\\\ = {\rm{2}}{\rm{.88}} \times {\rm{1}}{{\rm{0}}^{ - 7}}{\rm{C}}\\\end{array}$

Calculate the area of the sphere.

Write the expression for surface area.

$A = {\kern 1pt} 4\pi {R^2}$

Here, $R$ is the radius of conducting sphere.

Substitute $10{\rm{ cm}}$for $R$.

$\begin{array}{c}\\A = {\rm{4}}\pi {\left( {{\rm{10 cm}}} \right)^2}\\\\ = {\rm{4}}\pi {\left( {10{\rm{ cm}}\left( {\frac{{1{\rm{ m}}}}{{100{\rm{ cm}}}}} \right)} \right)^2}\\\\ = 0.1257{\rm{ }}{{\rm{m}}^2}\\\end{array}$

Calculate the surface charge density on the sphere.

Write the expression for surface charge density.

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

Here, $q$ is the charge on the conducting sphere and $A$ is the surface area of the sphere.

Substitute $2.88 \times {10^{ - 7}}{\rm{ C}}$ for $q$ and $0.1257{\rm{ }}{{\rm{m}}^2}$for $A$.

$\begin{array}{c}\\\sigma = \frac{{2.88 \times {{10}^{ - 7}}{\rm{ C}}}}{{0.1257{\rm{ }}{{\rm{m}}^2}{\rm{ }}}}\\\\{\rm{ }} = 2.29 \times {10^{ - 6}}{\rm{ C/}}{{\rm{m}}^2}\\\\{\rm{ }} = \left( {2.29 \times {{10}^{ - 6}}{\rm{ C/}}{{\rm{m}}^2}} \right)\left( {\frac{{1{\rm{ \mu C/}}{{\rm{m}}^2}}}{{{{10}^{ - 6}}{\rm{ C/}}{{\rm{m}}^2}}}} \right)\\\\{\rm{ }} = 2.29{\rm{ \mu C/}}{{\rm{m}}^{\rm{2}}}\\\end{array}$

(b)

Write the expression for the capacitance of the isolated spherical capacitor.

$C = 4\pi {\varepsilon _0}R$

Here, ${\varepsilon _0}$is the absolute permittivity and $R$ is the radius of the sphere.

Substitute $10{\rm{ cm}}$for $R$ and $8.854 \times {10^{ - 12}}{\rm{ F/m}}$ for${\varepsilon _0}$.

$\begin{array}{l}\\{\rm{C = 4}}\pi \left( {{\rm{8}}{\rm{.854}} \times {\rm{1}}{{\rm{0}}^{ - 12}}{\rm{ F/m}}} \right)\left( {10{\rm{ cm}}} \right)\\\\{\rm{C = 4}}\pi \left( {{\rm{8}}{\rm{.854}} \times {\rm{1}}{{\rm{0}}^{ - 12}}{\rm{ F/m}}} \right)\left( {10{\rm{ cm }}\left( {\frac{{1{\rm{ m}}}}{{100{\rm{ cm}}}}} \right)} \right)\\\\{\rm{ }} = \left( {{\rm{11}}{\rm{.1}} \times {\rm{1}}{{\rm{0}}^{ - 12}}{\rm{ F}}} \right)\left( {\frac{{1{0^{12}}{\rm{ pF}}}}{{1{\rm{ F}}}}} \right)\\\\{\rm{ }} = 11.1{\rm{ pF}}\\\end{array}$

Ans: Part a

The surface charge density on the sphere is $2.29{\rm{ \mu C/}}{{\rm{m}}^{\rm{2}}}$.

Part b

The capacitance of isolated sphere is $11.1{\rm{ pF}}$.