Transcript Electric Potential Difference Or Voltage

Electric Potential Difference
Or Voltage
Forces and Fields 8
Consider a mass positioned in a
gravitational field
Work  EP
Ep  mgh
3.00m
W  Fapp d
1.0kg
Ep  19.6J
W  9.81N (3.00m)
 29.4 J
2.00m
Fg
Fup
 A mass above a reference level (ground) possesses gravitational potential energy.
 Work must be done against the force of gravity to increase Ep. (Lifting the object
further off the ground increases Ep)
 Between the two points in the gravitational field above the ground,(2m and 5m) the
difference in Ep can be determined.
 Consider a charge placed within an electric field
Electric Potential Energy
Work  Ep
Fdown
Ep greater
-
Ep
d
E p lesser
-
Fup
Fe
Fe
-
E p lesser
Ep
d
-
E
Ep greater
E
+ charge
- charge
• moving the – charge away from the positive charge requires work which
increases electric potential energy and moving the – charge toward the
negative charge also does the same
•Electrical Ep changes when work is done in
moving a charge from one point to another
within an electric field
Ep lesser
Ep greater
Fapplied
E
-q1
+
+
Fe
d
Work  Fd  E p  increases E p
 To increase Ep between two unlike charges, move them farther
apart.
+
Low Ep
d
+
d
Higher Ep
•To increase electric Ep between two like
charges, move them closer together
d
+
d
+
Low Ep
+
+
Higher Ep
Electrical Potential Difference or Voltage (V)
 the ratio of the change in energy (work, potential or kinetic) per
unit charge (1.00 C) within an electrical field
Eg) two oppositely charged plates
E
-
uniform
-
+
+
+
+
High Ep
low Ep
+
+
-
+
-
+
-
+
Ep
+
-
 Between the two points that a positive charge has moved within the
electric field, there exists a voltage
Electric potential difference or voltage is the change in
potential energy per unit charge.
Equation:
In Joules
E
V
In Coulombs
q
 units are in J / C which is equivalent to volts
 The ratio of E/q is called a volt and is named after Alex Volta
who invented the battery
Examples:
 A 12.0 volt battery supplies 12.0 joules of energy to move a 1.00
C charge through a circuit.
 The voltage of 1.00 V between two points in an electric field
means that 1.00 J of energy is needed to move a 1.00 C charge
between the two points in the field.
1.00 V
+
-
+
-
+
+
+
Fe
-
+
-
Ek  0
+
+
-
Work  Fe d
-
-
 1.00 J  E p  1.00 J -
E 1.00 J
V

 1.00V
q 1.00C
Example:
 A charge of -2.60 X 10-3uC is moved from a positive plate to
a negative plate by doing 1.17 X 10-5 J of work. What is the
potential difference between the plates?
W  1.17 105 J
+
-
+
-
q  2.60 109 C
-
V
+
+
+
Fe
-
-
-
+
+
+
-
-
Fapplied
-
E
V
q
1.17  10 5 J

2.60  10 9 C
 4.50  103V
2nd Equation for Voltage
 The potential difference or voltage between two oppositely
charged plates is determined by the magnitude of the uniform
electric field, E, and the distance between the plates.
 Equation for voltage between oppositely charged plates:
E
V
q
Work, Ep
or Ek, energy
Fe
W  Fe d but E 
and Fe  Eq
q
So W  Eqd
Eqd
So V 
q
V
V  Ed and E 
d
Example:
 Determine the electric field strength between 2 oppositely
charged plates 25 cm apart to which 120 V is supplied.
0.25 m
-
+
+
+
+
E
-
+
-
+
-
+
-
+
V = 120 v
V  Ed
V
E
d
120V
E
0.250m
 480 V / m