# Electrical Potential

We know that gravity is a conservative force. Any work done by gravity or
against gravity is independent of the path taken. Work is measured simply as the change in gravitational potential energy. *D**U = mg**D*** h**. Similarly, an electric field is also a conservative force. The work required to move a
particle against a field depends only on the initial and final positions and is independent of the path taken. We say that:

where ** W** is the work,

**is the Field Strength,**

*E***is the charge, and**

*q***is the direct distance between two points. This work is equal to the Electric Potential Energy between two points. There are some things to note:**

*d*- Electrical Potential is independent of charge.
- This equation is only valid if the field is uniform.
- The sign needs to be considered in light of what you are doing. If you are moving a positive charge to an area of higher potential, you are doing work and increasing the charge's potential energy. If you move it to an area of lower potential (towards the negative) you are decreasing its potential energy. The opposite is true for a negative charge.
- Electrical potential energy is mechanical form of energy. That is to say, in the absence of friction or radiation, total mechanical energy, including electrical potential energy, will be conserved.

Equipotential
Surface: A circle (or locus of points) of radius "r" around a
charge where the field strength is the same. On the diagram, the equipotential lines are the dotted lines around the charge: Work is done by or against the field only when a charge is moved from one equipotential
point to another. This is analogous to moving one

from level of gravitational potential energy to another.

Electric Potential
is the measure of the size (magnitude) of electric potential energy per unit
charge at a particular location in the field. The work done, per unit charge, in moving from one equipotential surface to another equipotential surface is a measure of the Potential Difference, or Voltage. Voltage (V) is the potential difference (V_{A} - V_{B}) in moving from
point A to Point B.

Units? Of course we have units. The units are Joules/Coulomb, or Volts. 1 V = 1 J/C.

Let *r* = the distance between V_{A} and V_{B}. Then
from above:

or

The trick to the whole thing is finding surfaces that have equal potentials. We can define the units of electric field strength in terms of the potential difference between 2 points. If we pick one of those points at infinity, we can assign its potential to be zero.

Now, let's consider two parallel plates, one with a positive charge and one
with a negative charge. **d** represents the distance between the plates. If a charge *q* with mass *m* enters the field with a velocity *v*, it will be subject to
the same kinematics laws of motion as particles in a gravitational field. The force on the charge can be given in

terms of Newton's second law, or in terms of the Electric Field. By convention, a positive charge that enters an electric field will lose potential energy as it moves in the direction of the field.
It will gain kinetic energy.

Another term, *Grounding* refers to the connection of a
charged object to an object that has an electrical potential energy of zero.

Another unit of electrical energy is the Electron-Volt (eV). By definition, this is the energy required to move one electron through one volt of potential difference. 1eV = 1.6 x 10 ^{-19} Joules.

Charge flows when there is a difference in potential, much like water flows down hill. A constant flow of charge, called current, requires a constant potential difference. Using a continuous power source, such as a chemical battery or an electric generator can provide this potential difference. We'll talk more about current flow in another lesson.

Here is an example:

What is the speed of (a) a 900 eV, and (b) a 5.3keV, electron?

Solution: a) 900 eV = KE

900eV * 1.6x10^{-19}J/eV = 1/2
(9.1x10^{-31}kg)(v^{2})

v= 1.77x10^{7} m/s

b) 5300eV=KE

5300eV * 1.6x 10 ^{-19}J/eV = 1/2 (9.1x 10
^{-31}kg)(v^{2})

v=4.3x 10^{7} m/s

For further study and enlightenment, try:

http://www.physicsclassroom.com/Class/circuits/u9l1a.cfm

http://www.physicsclassroom.com/Class/circuits/U9L1b.cfm

http://www.physicsclassroom.com/Class/circuits/U9L1c.cfm

www.mhhe.com/physsci/physical/jones/student/olc/chapterindex17.htm

www.people.vcu.edu/~rgowdy/mod/083/imp.htm

http://physics.bu.edu/~duffy/PY106/Potential.html

For Practice Problems, Try:

*Giancoli Multiple Choice Practice Questions (Go ahead - try a
few.)*