### Section 2

```

As you lift an object off the ground, you are
increasing its potential energy
Same is for electric potential
◦ Electric potential (ΔV)
 Work done moving a test charge in an electric field by
dividing the magnitude of the test charge
 ΔV = W / q (Work is the Potential energy need to
remove the charge over some distance = joules)
◦ Measured in joules /coulomb (J/C = Volt (V))



Work is required to push a charged particle against the
electric field of a charged body.
EPE is the energy a charge particle possesses because of its
location in an electric field.
If the particle is released it will accelerate away turning
the EPE into kinetic energy.


If you apply 150 J of work to move a positive
charge of 3.5 x 10-6 C from a negative plate,
what is the electric potential difference?
Known
◦ Work on q = 150 J
◦ q=3.5 x 10-6 C

Unkown
◦ ΔV
◦ ΔV = W / q = 150 J / 3.5 x 10-6 C
◦ =

Electric Potential
◦ Smaller when two unlike charges are closer
together
◦ Larger when two like charges are


Uniform electric force and field made by
placing 2 large conducting plates parallel to
each other
◦ Direction is from + plate to –plate
Potential difference, ΔV, between 2 points a distance
(d) apart, in a uniform field (E)

ΔV = Ed


2 Parallel plates are given opposite charges.
A voltmeter measures the EPD to be 60.0 V.
The plates are 3.0 cm apart. What is the
magnitude of the electric field between them?
Known
◦ ΔV = 60.0 V
◦ D = 0.030 m

Unkown
◦ E = ???



E=V/d
= 60.0 V / 0.030 m
= 2.0 x 103 N/C

Storing energy in an electric field
◦ Leyden Jar
◦ Developed by Dutch physicist Pieter Van
Musschenbroek


Used by Ben Franklin to store charges from
lightning
Version is still used today: Capacitor


Ratio of charge stored to
electric potential difference:
called Capacitance, (C)
Capacitor designed to store
electric charges and energy
◦ Made of two conductors separated
by an insulator
◦ Capacitance = charge / electric
potential difference
◦ C = q / ΔV
◦ Measured in Coulomb per volt (C/V)


A sphere has an eletric potential difference
between it and Earth of 60.0 V when it has
been charged to 3.0 x 10-6 C. What is the
capacitance?
Known
◦ V = 60.0 V
◦ q = 3.0 x 10-6

Unknown
◦ C = ???




C = q / ΔV
= 3.0 x 10-6 / 60.0 V
= 0.00000005 F
= 0.05 µF

Examples: crank/shake flashlight, computer
keyboards, flashes in cameras, electronics.
```