Let's go back to junior physics for a second

Let's go back to junior physics for a second :)

• Let's go back to junior physics for a second :)

• What is gravitational potential energy?

• Energy that depends on an object's mass and its position relative to some point
• i.e. To calculate someone's potential energy relative to the surface of the Earth you'd need mass, g and height above the surface

The idea of electric potential energy is similar to that of gravitational potential energy

• The idea of electric potential energy is similar to that of gravitational potential energy

• Electric potential energy for a charge is calculated based on the magnitude of the charge and its position relative to some point

Let's say you have an electric field of magnitude 4500 N/C pointing toward the right

• Let's say you have an electric field of magnitude 4500 N/C pointing toward the right

• If you place a proton in that field, what is the magnitude and direction of the force acting on that proton?

• F=qE=7.2 x 10-16 N
• Right

So since the force acting on the proton is toward the right, it will accelerate toward the right

• So since the force acting on the proton is toward the right, it will accelerate toward the right

• What will happen to the proton's kinetic energy and electric potential energy?

• Kinetic Energy will increase
• EPE will decrease (conservation of energy)

Let's say we've got two charged plates that are separated by a small distance (this is a capacitor)

• Let's say we've got two charged plates that are separated by a small distance (this is a capacitor)

• The E-field points from left to right

A proton between these two plates would move towards the negative plate (right)

• A proton between these two plates would move towards the negative plate (right)

• An electron between these two plates would move towards the positive plate (left)

A proton has the highest potential energy when it's near the positive plate

• A proton has the highest potential energy when it's near the positive plate

• An electron has the highest potential energy when it's near the negative plate

By convention, the positive plate is at a higher potential than the negative plate

• By convention, the positive plate is at a higher potential than the negative plate

• Positively charged objects move from higher potential to lower potential (i.e. towards negative plate)
• Negatively charged objects move from lower potential to higher potential (i.e. towards positive plate)

Electric Potential, V, is the potential energy per unit charge

• Electric Potential, V, is the potential energy per unit charge

• Unit is Volts (1 V= 1J/1 C)

• If a point charge, q, has an electric potential energy at some point a, then the electric potential is

• V= PE/q

The change in potential energy of a charge, q, when moved between two points a and b

• The change in potential energy of a charge, q, when moved between two points a and b

• ΔPE = PEb-PEa=qVba

An electron in a television set is accelerated from rest through a potential difference Vba=+5000 V

• An electron in a television set is accelerated from rest through a potential difference Vba=+5000 V

• What is the change in PE of the electron?
• What is the speed of the electron as a result of the acceleration?
• Repeat for a proton that accelerates through a potential difference of -5000 V

ΔPE = Peb-PEa=qVba

• ΔPE = Peb-PEa=qVba

• ΔPE = qVba=(-1.6 x 10-19 C)(5000 V)

• ΔPE = -8 x 10-16 J

• Potential Energy was lost!

Conservation of Energy!

• Conservation of Energy!

• The amount of PE lost, must be equal to the amount of KE gained!

• KE= 8 x 10-16 J=0.5mv2

• V=4.2 x 107 m/s

ΔPE = qVba=(1.6 x 10-19 C)(-5000 V)

• ΔPE = qVba=(1.6 x 10-19 C)(-5000 V)

• ΔPE = -8 x 10-16 J (Same as electron)

• Velocity is less because speed is greater

• V=9.8 x 105 m/s

Since potential energy is always measured relative to some other point, only differences in potential energy are measurable

• Since potential energy is always measured relative to some other point, only differences in potential energy are measurable

• Potential Difference is also known as voltage

In order to move a charge between two points a and b, the electric force must do work on the charge

• In order to move a charge between two points a and b, the electric force must do work on the charge

• Vab=Va-Vb= -Wba/q

• The potential difference between two points a and b is equal to the negative of the work done by the electric force to move the charge from point b to point a, divided by the charge

How much work is needed to move a proton from a point with a potential of +100 V to a point where it is -50 V?

• How much work is needed to move a proton from a point with a potential of +100 V to a point where it is -50 V?

We're moving the proton from +100 V to -50 V

• We're moving the proton from +100 V to -50 V

• Therefore point A is +100 V, point B is -50 V

• We're looking for the work done by the field

• -Wba= qVab=q(Va-Vb)

• -Wba= (1.6 x 10-19 C)(100V -(-50V))

• Wba= -2.4 x 10-17 J

For two parallel plates, the relationship between electric field and electric potential is below

• For two parallel plates, the relationship between electric field and electric potential is below

• E=Vba/d

• d is the distance between the plates

The electron volt is another unit for energy

• The electron volt is another unit for energy

• 1 ev= 1.6 x 10-19 J

• Problem: A proton has 2 MeV of kinetic energy, how fast is it moving?

• 2 x106 eV= 3.2 x 10-13 J= 0.5mv2

• V= 1.96 x 107 m/s

Equipotential lines are used to represent electric potential

• Equipotential lines are used to represent electric potential

• Equipotential lines are always perpendicular to electric field lines

Equipotential lines (green) are perpendicular to the electric field lines (red)

• Equipotential lines (green) are perpendicular to the electric field lines (red)

The electric potential at a distance r from a single point charge q is : V=kQ/r

• The electric potential at a distance r from a single point charge q is : V=kQ/r

• Potential is zero at infinity

• The potential near a positive charge is large and decreases toward zero at large distances

The potential near a negative charge is negative and increases toward zero at large distances

• The potential near a negative charge is negative and increases toward zero at large distances

What minimum work is required by an external force to bring a charge q = 3.00 microC from a great distance away to a point 0.500 m from a charge Q= 20.0 microC?

• What minimum work is required by an external force to bring a charge q = 3.00 microC from a great distance away to a point 0.500 m from a charge Q= 20.0 microC?

Basically, we're taking the charge q from a place of zero potential, to a place of nonzero potential

• Basically, we're taking the charge q from a place of zero potential, to a place of nonzero potential

• Use our trusty equation:Vab=Va-Vb= -Wba/q

The charge is coming from infinity, so Va=0

• The charge is coming from infinity, so Va=0

• What is Vb?

• Vb=KQ/r=(9x109 Nm2/C2)(20x10-6C)/0.500m
• Vb= 360,000 V

• Wba= -q(Va-Vb)=-(3.00x10-6C)(0-360000V)

• W= 1.08 J

Electric fields are vectors, but electric potential is a scalar!

• Electric fields are vectors, but electric potential is a scalar!

• When determining the electric potential at a point you can just add the electric potential from each charge, just be sure to include the correct sign of the charge when calculating potential

Calculate the electric field at a point midway between a -0.5 microC charge and a -0.8 microC charge that are separated by 0.50 m.

• Calculate the electric field at a point midway between a -0.5 microC charge and a -0.8 microC charge that are separated by 0.50 m.

• For the -0.5 microC charge, E= 72000 N/C left

• For the -0.8 microC charge, E= 115,200 N/C right

• Therefore E is 43200 N/C right

Calculate the electric potential at a point midway between a -0.5 microC charge and a -0.8 microC charge that are separated by 0.50 m.

• Calculate the electric potential at a point midway between a -0.5 microC charge and a -0.8 microC charge that are separated by 0.50 m.

• For the -0.5 microC charge,

• V=kQ/r= (9x109 Nm2/C2)(-0.5 x 10-6 C)/0.25m

• V= -18000 N/C

For the -0.8 microC charge,

• For the -0.8 microC charge,

• V=kQ/r= (9x109 Nm2/C2)(-0.8 x 10-6 C)/0.25m

• V= -28800 V

• Total V= -46800 V

• This is much easier! No directions...just make sure you include the sign!

A capacitor stores electric charge and consists of two conducting objects that are placed next to each other but not touching

• A capacitor stores electric charge and consists of two conducting objects that are placed next to each other but not touching

If a voltage is applied to a capacitor (i.e. connected to a battery), then it becomes charged

• If a voltage is applied to a capacitor (i.e. connected to a battery), then it becomes charged

• Amount of charge for each plate:

• C= Capacitance of the capacitor (different for each capacitor)
• Unit for C is farad (F)

A= Area of plates

• A= Area of plates

• If A increases, C increases

• d= distance between the plates

• If d increases, C decreases

• ε0 = 8.85 x 10-12 C2/Nm2

• (This is the permitivity of free space)

A charged capacitor stores electric energy

• A charged capacitor stores electric energy

A 7.7 µF capacitor is charged by a 125 V battery and then is disconnected from the battery. When this capacitor (C1) is connected to a second, uncharged capacitor (C2), the voltage on the first drops to 15 V. What is the value of C2? (Charge is conserved)

• A 7.7 µF capacitor is charged by a 125 V battery and then is disconnected from the battery. When this capacitor (C1) is connected to a second, uncharged capacitor (C2), the voltage on the first drops to 15 V. What is the value of C2? (Charge is conserved)

For the first capacitor:

• For the first capacitor:

• When the capacitors are connected, the voltage on the first one is 15 V. That means the new charge on C1 is:

What happens to the rest of the charge?

• What happens to the rest of the charge?

• It must be on capacitor 2 because charge is conserved
• Since the two capacitors are connected, the voltage for the second one must also be 15 V

Capacitors can be connected in series or parallel

• Capacitors can be connected in series or parallel

• When capacitors are connected in parallel, the equivalent capacitance is the sum

• The voltage across each capacitor is the same

If the capacitors are connected in series, the equivalent capacitance is given by the following expression

• If the capacitors are connected in series, the equivalent capacitance is given by the following expression

For capacitors in series, the total voltage must equal the sum of the voltages across each capacitor

• For capacitors in series, the total voltage must equal the sum of the voltages across each capacitor

• The charge on each capacitor is the same as the charge on the equivalent capacitor for capacitors in series

What is the equivalent capacitance for this combination of capacitors?

• What is the equivalent capacitance for this combination of capacitors?

• C2 and C3 are connected in parallel

• Combine them into one capacitor

• C23=C2 + C3 = 35 µF

C23 and C1 are connected in series

• C23 and C1 are connected in series

How much charge is stored on each capacitor?

• How much charge is stored on each capacitor?

• Q=CV
• V1= 50 V (this is the voltage across C1)

C1 and C23 are connected in series, therefore the charge on C23 is the same as the charge on C1

• C1 and C23 are connected in series, therefore the charge on C23 is the same as the charge on C1

C2 and C3 are connected in parallel, therefore:

• C2 and C3 are connected in parallel, therefore:

The charge on C2 is:

• The charge on C2 is:

• The charge on C3 is: