Question

Three equal point charges, each with charge 1.10 $\mu C$ , are placed at the vertices of an equilateral triangle whose sides are of length 0.350 . What is the electric potential energy of the system? (Take as zero the potential energy of the three charges when they are infinitely far apart.)

Use $epsilon_0$ = 8.85×10⁻¹² for the permittivity of free space.

Answer 1

Concepts and reason

The given problem involves the electric potential energy due to the system of charges. The energy is a scalar quantity. So, the electric potential energy between the two charges can be added directly.

Let the charges that are placed at the vertices of the equilateral triangle be${q_1}{\rm{, }}{q_2}{\rm{ and }}{q_3}$.

First, find the electric potential energies between set of charges${q_1}{\rm{ and }}{q_2}$,${q_1}{\rm{ and }}{q_3}$, and ${q_2}{\rm{ and }}{q_3}$. Then, add all the individual electric potential energies in order to get the total electric potential energy of the given system of charges.

Fundamentals

The electric potential energy is the work done in moving a charge in the electric field from infinity to a certain distance.

The electric potential energy U between the charges ${q_1}{\rm{ and }}{q_2}$ separated by a distance r can be expressed as follows:

$U = k\frac{{{q_1}{q_2}}}{r}$

Here, $k$ is the constant with value $9 \times {10^9}{\rm{ N}} \cdot {{\rm{m}}^2}{\rm{/}}{{\rm{C}}^2}$.

For an equilateral triangle, all sides are same and all the angles are also the same.

Given that the three charges, which are placed at the vertices of the equilateral triangle are same. Let the charge be equal to q.

${q_1} = {q_2} = {q_3} = q$

All the charges placed at the vertices of the equilateral triangle separated by the same distance r.

The electric potential due to the charges ${q_1}$ and ${q_2}$ separated by the distance r is,

${U_{12}} = k\frac{{{q_1}{q_2}}}{r}$

As the charges are of same magnitude, replace ${q_1}{\rm{ and }}{q_2}$with q.

$\begin{array}{l}\\{U_{12}} = k\frac{{\left( q \right)\left( q \right)}}{r}\\\\{U_{12}} = k\frac{{{q^2}}}{r}\\\end{array}$

The electric potential due to the charges${q_1}{\rm{ and }}{q_3}$ separated by the distance r is,

${U_{13}} = k\frac{{{q_1}{q_3}}}{r}$

As the charges are of same magnitude, replace ${q_1}{\rm{ and }}{q_3}$with q.

$\begin{array}{l}\\{U_{13}} = k\frac{{\left( q \right)\left( q \right)}}{r}\\\\{U_{13}} = k\frac{{{q^2}}}{r}\\\end{array}$

The electric potential due to the charges${q_2}{\rm{ and }}{q_3}$ separated by the distance r is,

${U_{23}} = k\frac{{{q_2}{q_3}}}{r}$

As the charges taken are same, replace ${q_2}{\rm{ and }}{q_3}$with q.

$\begin{array}{l}\\{U_{23}} = k\frac{{\left( q \right)\left( q \right)}}{r}\\\\{U_{23}} = k\frac{{{q^2}}}{r}\\\end{array}$

The total electric potential energy of the given system of three charges is the sum of electric potential energies of${q_1}{\rm{ and }}{q_2}$, ${q_1}{\rm{ and }}{q_3}$, and ${q_2}{\rm{ and }}{q_3}$. Thus,

$U = {U_{12}} + {U_{13}} + {U_{23}}$

Replace $k\frac{{{q^2}}}{r}$for all${U_{12}}$, ${U_{13}}$, and${U_{23}}$.

$\begin{array}{c}\\U = k\frac{{{q^2}}}{r} + k\frac{{{q^2}}}{r} + k\frac{{{q^2}}}{r}\\\\U = 3k\frac{{{q^2}}}{r}\\\end{array}$

Substitute $9 \times {10^9}{\rm{ N}} \cdot {{\rm{m}}^2}{\rm{/}}{{\rm{C}}^2}$for$k$, $1.10{\rm{ }}\mu {\rm{C}}$for$q$, and $0.350{\rm{ m}}$for$r$.

$\begin{array}{c}\\U = 3\left( {9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^2}{\rm{/}}{{\rm{C}}^2}} \right)\frac{{{{\left( {1.10{\rm{ }}\mu {\rm{C}}\left( {\frac{{{{10}^{ - 6}}{\rm{ C}}}}{{1{\rm{ }}\mu {\rm{C}}}}} \right)} \right)}^2}}}{{0.350{\rm{ m}}}}\\\\ = 9.33 \times {10^{ - 2}}{\rm{ J}}\\\end{array}$

Ans:

Thus, the electric potential energy of the given system of three charges that are placed at the vertices of an equilateral triangle is $9.33 \times {10^{ - 2}}{\rm{ J}}$.