Derive an expression for electric potential energy of a system of three points charges

Derive an expression for potential energy of a system of three point charges

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In this article, we are going to derive an expression for potential energy of a system of three points charges. So keep reading till end..

DERIVATION FOR THE POTENTIAL ENERGY OF A SYSTEM OF THREE POINTS CHARGES

Let’s consider a system of three points charges q_1, q_2 and q_3 having position vector r_1, r_2 and r_3 respectively from the origin as shown in following figure.

Derive an expression for electric potential energy of a system of three points charges
System of three point charges

If we bring charge q_1 first from infinity to position r_1 then there is no work done required to do so, it is because when we bring charge q_1 from infinity to position r_1 then at that position there is no any source which can produce electric field. If there is no electric field then there is no any opposing force. Hence work done is zero.

    \[\therefore W_1=0\]

But when we bring charge q_2 from infinity to the position r_2 then in this, we have to do work done because, here a opposing field is present due to charge q_1. So we have to do work done in against the electric field produced by the first electric charge q_1.
The work done in bringing charge q_2 from infinity to the position r_2 is-

    \begin{align*} W_2&=q_2V\left(r_2\right)\\ W_2&=\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r_{12}} \end{align*}

Where r_{12} is the position vector between charge q_1 and q_2 and V is the electric potential due to charge q_1 at position vector r_{12}.
Now, charges q_1 and q_2 will produce a electric potential at any point say P. Think that point P denotes the position of charge q_3. The position vector between charge q_1 and q_3 will be r_{13} and in between charge q_2 and q_3 will be r_{23}.

Now the electric potential due to the charge q_1 and q_2 at point P is given as-

    \[V_{1,2}=\frac{1}{4\pi\epsilon_0}\left(\frac{q_1}{r_{13}}+\frac{q_2}{r_{23}}\right)\]

So the work done in bringing charge q_3 from infinity to the position r_3 is-

    \begin{align*} W_3&=q_3V_{1,2}\left(r_3\right)\\ &=\frac{1}{4\pi\epsilon_0}\left(\frac{q_1q_3}{r_{13}}+\frac{q_2q_3}{r_{23}}\right) \end{align*}

The total work done in assembling the charges at the given location is equal the total potential energy of the system and According to the superposition principle, this total potential energy can be obtained by adding the work done of individual charges.

    \begin{align*} U&=W_1+W_2+W_3\\ &=0+\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r_{12}}+\frac{1}{4\pi\epsilon_0}\left(\frac{q_1q_3}{r_{13}}+\frac{q_2q_3}{r_{23}}\right) \end{align*}

    \[\boxed{U=\frac{1}{4\pi\epsilon_0}\left(\frac{q_1q_2}{r_{12}}+\frac{q_1q_3}{r_{13}}+\frac{q_2q_3}{r_{23}}\right)}\]

This result can also be expressed in the form of summation as follows-

    \[U=\left[\frac{1}{4\pi\epsilon_0}\sum_{i=1}^3\sum_{i=1_{j\neq i}}^3\frac{q_iq_j}{r_{ij}}\right]\]

If we want to obtain the value of electric potential energy of a system of N point charges then we also obtained it. The value of electric potential energy due to a system of N point charges is equal to the total amount of work done in assembling all the charges to the given position from infinity.

    \begin{align*} U&=\left[\frac{1}{4\pi\epsilon_0}\sum_{i=1}^N\sum_{i=1_{j\neq 1}}^Nq_iV_j\right]\\ \text{where}\; V_j&=\sum_{i=1_{j\neq i}}^N\frac{1}{4\pi\epsilon_0}\frac{q_i}{r_{ij}} \end{align*}

Watch this video for more reference:

We know that electrostatic force is conservative in nature, so the value of U is independent of the manner in which the configuration of charge is assembled.
The SI unit of electric potential energy is joule (J) and it’s another unit is electron volt (eV)

    \begin{align*} 1eV&=1.6\times10^{-19}\times1V\\ &=1.6\times10^{-19}J \end{align*}

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