Electrical Potential Energy

A form of Potential Energy that depends on the configuration of charges inside a System.

Definition §

Using the Electrical Potential of a singular Point Charge, we can combine it with another point charge to get the potential energy.

Assume we have two point charges, $q_1$ and $q_2$ respectively, separated by distance $r$

Formula §

$$U_E=k_e\times \frac{q_1\times q_2}{r}$$

• $U_E$ = Electropotential Energy (in $J$) of the system
• $q$ = Charge (of $A$ and $B$)
• $r$ = Distance between A and B.

Graph §

We can use a graph of this to understand concepts better. Note that $U_E\propto \frac{1}{r}$

Derivation §

Assume we have an isolated system of two Charges, $q_1$ and $q_2$. This system has no external forces, and the only internal force is the Electrostatic Force between the charges.

The formula for electrostatic force is derived from Coulomb’s Law:

The electrostatic force ($F_{elec}$) is only zero in our proposed system if the two charges are infinitely far apart, regardless of whether they are like or unlike charges.

So, our ‘zero’ point for electrical potential energy is best suited to be when the charges are infinitely far apart, i.e. $r=\infty$.

If we define $U_{elec}=0$ when $r=\infty$, then any other configuration of the system results in a non-zero potential energy.

Unlike Charges §

Let $q_1=+Q$ and $q_2=-Q$

In order to keep $\sum F=0$, we have an external force that acts anti-parallel to the displacement, but with the same magnitude as the electric force, i.e. ${\vec{F}}_{ext}=-{\vec{F}}_{elec}$

In order to move the system from C1 to C2, we need to apply some work:

The formula for Work done is:

Here, we want to calculate the work done by the external force to change the distance from $\infty$ to $r_f$.

The displacement, $\vec{x}$, can be expressed as a change in r, $d\vec{r}$

Because the force is non constant, we need to integrate the total work.

$$W=\vec{F}\cdot \vec{x}=\int {\vec{F}}_{ext}\cdot \text{d}\vec{x}=-\int |F_{ext}|\text{d}x\text{( as vectors are anti-parallel)}$$

$$W_{ext}=-{\int }_{r_i}^{r_f}|F_{ext}|\text{d}r$$

$$W_{ext}=-k_e{Q}^2{\int }_{r_i}^{r_f}\frac{1}{r^2}\text{d}r$$

$$W_{ext}=-k_e{Q}^2[\frac{1}{r}{]}_{r_i}^{r_f}=k_e{Q}^2[\frac{1}{r_f}-\frac{1}{r_i}]$$

At the same time, we have a conservative, internal force acting on the charge - the electrostatic force, $F_{elec}$. Because the charges are unlike, this operates similarly to Gravitational Force, and is parallel to the displacement.

$$W_c=\vec{F}\cdot \vec{x}=\int {\vec{F}}_{elec}\cdot \text{d}\vec{x}=\int |F_{elec}|\text{d}x\text{( as vectors are parallel)}$$

$$W_c={\int }_{r_i}^{r_f}|F_{elec}|\text{d}r$$

$$W_c=k_e{Q}^2{\int }_{r_i}^{r_f}\frac{1}{r^2}\text{d}r$$

$$W_c=k_e{Q}^2[\frac{1}{r}{]}_{r_i}^{r_f}=-k_e{Q}^2[\frac{1}{r_f}-\frac{1}{r_i}]$$

Also, this shows that:

$$\sum W=W_{ext}+W_c=0$$

And by the Work-Energy Theorem:

$$\sum W=0\to \Delta K=0$$

The change in potential energy, $U_{elec}$, comes directly from $W_c$, as it is a result of the work done by a conservative force:

$$U=-k_e{Q}^2(\frac{1}{r_f}-\frac{1}{r_i})=-\frac{k_e{Q}^2}{r}-0$$

Recall that $q_1=Q$ and $q_2=-Q$

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$$U_{elec}=\frac{k_e{q}_1{q}_2}{r}$$

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Like Charges §

Essentially, the same thing happens, only this time we have the force opposing the movement. In other words, because the electric force is anti-parallel to the displacement, we have the dot product being negative. Our external force is now towards the displacement:

Let $q_1=+Q$, $q_2=+Q$

In order to bring the system from config 1 to config 2, we essentially do the same thing as with a system of unlike charges:

$$W=\vec{F}\cdot \vec{x}=\int {\vec{F}}_{ext}\cdot \text{d}\vec{x}=\int |F_{ext}|\text{d}x(\text{vectors are parallel})$$

$$W={\int }_{r_i}^{r_f}|F_{ext}|\text{d}r$$

$$W=k_e{Q}^2{\int }_{r_i}^{r_f}\frac{1}{r^2}dr$$

$$W=k_e{Q}^2[\frac{-1}{r}{]}_{r_i}^{r_f}=-k_e{Q}^2[\frac{1}{r_f}-\frac{1}{r_i}]$$

And the associated conservative work done:

$$W_c=\vec{F}\cdot \vec{x}=\int {\vec{F}}_{elec}\cdot \text{d}\vec{x}=-\int |F_{elec}|\text{d}x(\text{vectors are anti-parallel})$$

$$W_c=-{\int }_{r_i}^{r_f}|F_{elec}|\text{d}r$$

$$W_c=-k_e{Q}^2{\int }_{r_i}^{r_f}\frac{1}{r^2}dr$$

$$W_c=-k_e{Q}^2[\frac{-1}{r}{]}_{r_i}^{r_f}=k_e{Q}^2[\frac{1}{r_f}-\frac{1}{r_i}]$$

We end up getting the same result:

$$U=k_e{Q}^2(\frac{1}{r_f}-\frac{1}{\infty })=\frac{k_e{Q}^2}{r_f}$$

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$$U_{elec}=\frac{k_e{q}_1{q}_2}{r}$$

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Change in $U$ §

$$\Delta U=U_f-U_i=k_e{q}_1{q}_2(\frac{1}{r_f}-\frac{1}{r_i})$$

$$\Delta U=W_{ext}=-W_c$$