Electrical Potential

The electrical potential is how much energy is needed to move a charge against an electric field. It can be thought of as the Energy version of the Electric Field. It is a relative concept, so we need to define a given charge/point to have 0 electrical potential, $V_i=0$.

Electrical potential vs Electrical Energy

Electrical potential, $V$, is not the same as Electrical Potential Energy, $U$! $U$ is a property of a System of charges, and does not exist in individual charges. If we have a system of charges, $q_1$ and $q_2$:

• $U_{sys}=\frac{k_e{q}_1{q}_2}{r}$
• $V_{q_1}=\frac{k_e{q}_1}{r}$

Signs!

$V$ of a particle is the same sign as the charge of the particle or the electric field experienced by that particle, $\vec{E}$. In a system of like charges, $U$ is always positive, and the electric field, $\vec{E}$ is positive, so $\frac{U}{q}=\frac{+ve}{+ve}=\text{positive}$
In a system of unlike charges, $U$ is always negative:

• If we look at the positive charge: It experiences an electric field associated with a negative charge, so $\text{sign}(\vec{E})=-ve$. $\frac{U}{q}=\frac{-ve}{+ve}=-ve=\text{sign}(\vec{E})$
• If we look at the negative charge, it experiences a positive electric field, so $\text{sign}(\vec{E})=+ve$. $\frac{U}{q}=\frac{-ve}{-ve}=+ve=\text{sign}(\vec{E})$

Definition §

Assume we have a charge, $+q$ that moves along a path ($L$) starting at point $i$ and ends at point $f$. If this path is inside a constant electric field, of strength $\vec{E}$, the electrical potential of $q$ is

Formula - 1 §

Change in Electric Potential from Electric Field

\Delta V_{i\to f}= - \int_{i}^f \vec{E} \cdot , d\vec{L}

##### Formula - 2 >[!formula] Change in Electric Potential from [Electrical Potential Energy](Electrical%20Potential%20Energy.md) > >$$ >\Delta V = \dfrac{\Delta U}{q} >$$ ##### Terms * $V$ = Electrical potential ( in $V$ or $J/C$ ) * $V_{i}, V_{f}$ = Initial, final electrical potential * $\vec{E}$ = [Electric Field](Electric%20Field.md) (in $N/C$ or $N/m$) * $\vec{L}$ = Path taken * $\Delta U$ = [Electrical Potential Energy](Electrical%20Potential%20Energy.md) of the system ### Superposition Because electrical potential is defined as a scalar, the total change in electrical potential is the sum of the changes in electrical potential caused by individual fields/particles. For a [Discrete Charge Distribution](Discrete%20Charge%20Distribution.md) of charges $q_{1}, q_{2}, \dots q_{n}$ at distances of $r_{1},r_{2},\dots r_{n}$ away from a given point, the electric potential is:

\sum \Delta V = k_{e}[\dfrac{q_{1}}{r_{1}} + \dfrac{q_{2}}{r_{2}} + \dots \dfrac{q_{n}}{r_{n}}]

$$Fora[ContinuousChargeDistribution](Continuous$$

\sum \Delta V = -\int_{i_{1}}^f \vec{E}{1} \cdot, d\vec{L} -\int{i_{2}}^f \vec{E}2 \cdot, d\vec{L} -\dots\int{i_{n}}^f \vec{E}_n \cdot, d\vec{L}

### Loops Since the electrical potential is derived as a [Path Integral](Path%20Integral), if $i$ = $f$, then:

\Delta V= - \int_{i}^f \vec{E} \cdot , d\vec{L} = 0

Therefore, any closed loop has $\Delta V$ = 0 ### Derivation Using the [derivation of electrical potential energy](Electrical%20Potential%20Energy.md#Derivation), we can generalise the energy to be irrespective of the other charge in the system, similar to how we generalise the [Electric Field](Electric%20Field.md) from the derivation of [Electrostatic Force](Electrostatic%20Force.md) to be irrespective of the other charge in the system.