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

>[!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 by the charge >* $\Delta U$ = [Electrical Potential Energy](Electrical%20Potential%20Energy.md) of the system

Formula - 2 §

Change in Electric Potential from Electrical Potential Energy

$$\Delta V=V_f-V_i=\frac{\Delta U}{q}$$

Note

• $\Delta V$ = Change in electrical potential ( in $V$ or $J/C$ )
• $V_i,V_f$ = Initial, final electrical potential
• $\Delta U$ = Electrical Potential Energy of the system
• $q$ = Charge of the particle (in $\text{C}$)

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 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:

$$\sum \Delta V=k_e[\frac{q_1}{r_1}+\frac{q_2}{r_2}+\dots \frac{q_n}{r_n}]$$

For a Continuous Charge Distribution, we need to integrate:

$$\sum \Delta V=-{\int }_{i_1}^f{\vec{E}}_1\cdot d\vec{L}-{\int }_{i_2}^f{\vec{E}}_2\cdot d\vec{L}-\cdots {\int }_{i_n}^f{\vec{E}}_n\cdot d\vec{L}$$

Loops §

Since the electrical potential is derived as a Path Integral, 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, we can generalise the energy to be irrespective of the other charge in the system, similar to how we generalise the Electric Field from the derivation of Electric Force to be irrespective of the other charge in the system.