Electric Potential

Motivation: Electric Potential §

• We wish to bring in a test charge $q$ from infinity to point $P$; $P$ is in the presence of an existing charge $Q$ in space
• $Q$’s electric field will try to repel $q$; we must do work to overcome the force exerted on $q$ and move $q$
• How is work defined?
  $$W={\int }_{\infty }^P\vec{F}\cdot \overrightarrow{dl}\begin{array}{rl}{F}_I& =\text{force needed to move q}\\ dl& =\text{differential path element}\end{array}$$
• All forces must have an equal and opposite reaction force; the force we exert to move $q$ must then be equal to $Q$’s electric field, which is ${\vec{F}}_e=q\vec{E}$
• Then, the force we exert is

$${\vec{F}}_I=-q\vec{E}$$

• ==Then, the work needed to move a point charge $q$ can be written as==

$$\begin{array}{r}W=-q_o{\int }_{\infty }^P\vec{E}\cdot \overrightarrow{dl}\end{array}$$

Procedure: Electric Potential to Move a Point Charge §

• Electric potential is the work done per positive unit test charge as $q$ approaches 0

$$\begin{array}{r}{V}_p=-{\int }_{\infty }^P\vec{E}\cdot \overrightarrow{dl}=\frac{kQ}{R}(V)\end{array}$$

• $V_p$ is always defined assuming a positive test charge
• The reference point is infinity as convention
• As electric fields are conservative, we obtain the same electric potential no matter what path taken to get to $P$; the simplest path would always be a straight line from $\infty$ to $P$
• ==If $Q>0$, then the positive test charge will be repelled and $V_p>0$, so we are doing work against the electric field
• ==If $Q<0$, then the positive test charge will be attracted and $V_p<0$, so the electric field is doing the work

Procedure: Electric Potential for Point and Charge Distributions §

• The general equation we found for electric potential for a point is very flexible

$$\begin{array}{rl}{V}_p& =\frac{kQ}{R}\\ & \\ d{V}_p& =\frac{kdQ}{R}\\ & \\ & =\frac{k\cdot \rho \cdot dV}{R}\end{array}$$

Definition: Electrostatic Potential Energy §

• Analogy: the gravitational potential at a point is the gravitational potential energy of a unit mass placed at that point
• The electric potential at any point in the electric field is the electric potential energy of a unit positive charge at that point
  • Electric potential is amount of work needed to bring point charge $q$ from infinity to $P$
  • Electric potential energy is the energy needed to move a charge against the electric field
• Def:

$$\begin{array}{r}\Delta U=-W_{a\to b}=-q_o{\int }_a^b\vec{E}\cdot \overrightarrow{dl}\end{array}$$

• Negative $dW$ means that work was done by the electric field
• Positive $dW$ means that work was done against the electric field

Differential Length Review

$$\begin{array}{rl}dl& =dx\hat{x}+dy\hat{y}+dz\hat{z}\text{Cartesian }\\ dl& =dr\hat{r}+r\cdot d\varphi \hat{\varphi }+dz\hat{z}\text{Cylindrical }\\ dl& =dr\hat{r}+r\cdot d\theta \hat{\theta }+\sin (\theta )\cdot d\varphi \hat{\varphi }\text{Spherical}\end{array}$$

Electric Potential of Charge Distribution §

• There are two methods of calculating the potential at $P$:

Conceptual Tricks §

Charge, Potential, E-field §

• Because $E=-\nabla V$, the electric field always points in the direction of decreasing potential
• Change of potential energy is $\Delta U=q\cdot \Delta V$
• If $\Delta U$ is negative, the electric field is doing the work and $W_{field}>0$, $W_{external}<0$
• If $\Delta U$ is positive, we (outside forces) are doing the work and $W_{field}<0$, $W_{external}>0$
• Change of potential energy is the work done by an external agent; the electric field does opposite work
• Electric field has to be perpendicular to equipotential lines or surfaces
• When the question asks for “calculate potential for all r”, make sure to consider various cases of r, such as r larger than radius, smaller than radius, etc.