Question

Two small spheres spaced 20.0 cm apart have equal charge.

 

How many excess electrons must be present on each sphere if the magnitude of the force of repulsion between them is 4.57 x 10^-21 N?

 

Number of Electrons = ?

Answer 1

Concepts and reason

Initially calculate the magnitude of charge on each small sphere using definition of electrostatic force between two charged spheres. Then use quantization of charge to find the excess number of electrons present on each sphere.

Fundamentals

The electrostatic force ${F_e}$ between two charges ${q_1}$ and ${q_2}$ separated by a distance r is given as follows:

${F_{\rm{e}}} = \frac{{k{q_1}{q_2}}}{{{r^2}}}$

Here, k is the coulomb’s constant.

The charge quantization is given as follows:

$q = ne$

Here, q is the charge on a particle, n is the number of electrons, and e is the charge on an electron.

The electrostatic force ${F_e}$ between two spheres separated by a distance r is given as follows:

${F_e} = k\frac{{{q_1}{q_2}}}{{{r^2}}}$

Here, ${q_1}$ is charge on one sphere, and ${q_2}$ is charge on another sphere.

Substitute $q$ for ${q_1}$ and ${q_2}.$

$\begin{array}{l}\\{F_e} = k\frac{{qq}}{{{r^2}}}\\\\{F_e} = k\frac{{{q^2}}}{{{r^2}}}\\\end{array}$

Rearrange the above equation for q.

$q = r\sqrt {\frac{{{F_e}}}{k}}$

Convert the units of distance r from cm to meters as follows:

$\begin{array}{c}\\r = 20\,{\rm{cm}}\left( {\frac{{1\,{\rm{m}}}}{{100\,{\rm{cm}}}}} \right)\\\\ = 0.2\,{\rm{m}}\\\end{array}$

Substitute 0.2 m for r, $9 \times {10^9}{\rm{ N}} \cdot {{\rm{m}}^2}/{{\rm{C}}^2}$ for k, and $4.57 \times {10^{ - 21}}{\rm{ N}}$ for ${F_{\rm{e}}}$ in the above equation $q = r\sqrt {\frac{{{F_e}}}{k}}$ and solve for q.

$\begin{array}{c}\\q = \left( {0.2\,{\rm{m}}} \right)\sqrt {\frac{{4.57 \times {{10}^{ - 21}}\,{\rm{N}}}}{{9 \times {{10}^9}\,{\rm{N}} \cdot {{\rm{m}}^2}/{{\rm{C}}^2}}}} \\\\ = 1.425 \times {10^{ - 16}}\,{\rm{C}}\\\end{array}$

Rearrange the expression $q = ne$ for n.

$n = \frac{q}{e}$

Substitute $1.425 \times {10^{ - 16}}\,{\rm{C}}$ for q, and $1.6 \times {10^{ - 19}}\,{\rm{C}}$ for e, in the above equation and solve for n.

$\begin{array}{c}\\n = \frac{{1.425 \times {{10}^{ - 16}}\,{\rm{C}}}}{{1.6 \times {{10}^{ - 19}}\,{\rm{C}}}}\\\\ = 891\,{\rm{electrons}}\\\end{array}$

Ans:

The excess number of electrons on each sphere is 891.