What unit measures charge, and what signs can be given to it?

The **Coulomb, C,** is the SI unit of charge and can have values that are positive (for current or protons) or negative (for electrons).

Since the AP Physics exam is concerned primarily with current (+) moving, Coulombs will primarily be positive values.

What are the rules for attraction of charges?

**Opposite charges** (+,-) are **attracted **towards each other.

The force of attraction varies proportionally with magnitude of charges.

What are the rules for repulsion of charges?

**Like charges** (+,+ or -,-) are **repulsed** away from each other.

The force of repulsion varies proportionally with magnitude of charges.

What must be true of the total charge in a closed system?

Total charge in a closed system will stay constant. This is the **law of conservation of charge.**

Describe the total charge before and after in the following closed system: 5 protons and 4 electrons collide with sufficient energy to create 4 neutrons and 1 proton.

Charge remains constant at +1.

Charge initially is: +5 + (-4) = +1.

In a closed system charge must be conserved. The final charge confirms this: 4(0) + (+1) = +1.

**conductance **of a material

**Conductance** is the ability of a material to transfer charge.

Conductance is generally higher in metals and lower in nonmetals.

a **conductor**

A **conductor **contains many electrical charges within the medium, and they are relatively moveable.

Common conductors are usually metals with high atomic weight, such as silver, copper, and gold.

Calcium contains 3x more moveable electric charges than Iron. Which is a better conductor?

Calcium has 3x the conductance of iron.

The more movable charges a material has, the higher its relative conductance.

**insulation**

**Insulation **is the degree to which a material is unable to transfer charge.

Insulation is generally higher in polymers, amorphous crystals, or ionic solids. Insulation is low in all other substances.

an **insulator**

An **insulator **has few free electrical charges within the medium, and those are difficult to move.

Common insulators are as glass, quartz, rubber and teflon.

Rubber has 10,000x more moveable electric charges than paraffin. Which is a better insulator?

Paraffin has 10,000x higher resistance (insulation) than rubber.

The fewer moveable charges a material has, the greater its relative resistance (insulation).

What is the relationship for the force between two charged particles separated at a given distance?

**Coulomb's Law**, defined as:

**F = Kq _{1}q_{2} / r^{2}**

Where:

K = Coulombic Constant in Nm^{2}/C^{2}

q_{1} and q_{2} = charges in C

r = distance between the charges in m

What does a positive value of F indicate, in Coulomb's Law for electrostatic force?

A **positive **value of F indicates that the charges are the same sign, and will experience a force of **repulsion**.

What does a negative value of F indicate, in Coulomb's Law for electrostatic force?

A **negative **value of F indicates that the charges are opposite in sign, and will experience a force of **attraction**.

What will be the change in force, if two positive charges separated by a distance r are now moved to a distance 2r apart?

The charges will now experience 1/4 original force.

From: F = Kq_{1}q_{2} / r^{2} the force is inversely proportional to the square of distance. Doubling distance will reduce magnitude of force to 1/4 its original value.

What will be the change in force, if a negative and positive charge at a distance r are replaced by equivalent but now both negative charges?

Force will change from negative (attraction) to positive (repulsion).

Magnitude will be the same, since the strength of the charges remains the same.

What concept and direction describes the path of motion that a positive test charge will travel, when put next to a stationary positive charge?

**Electrostatic field lines. **

By convention, all positive charges have field lines with arrows pointing outwards, since a positive test charge will be repulsed.

What concept and direction describes the path of motion that a positive test charge will travel, when put next to a stationary negative charge?

**Electrostatic field lines. **

By convention, all negative charges have field lines with arrows pointing inwards, since a positive test charge will be attracted.

Describe the motion of a positive test charge, if it's placed exactly at the midpoint between two equal positive stationary charges.

It will remain immobile.

The two positive charges repel the test positive charge equally, hence cancelling each other's force vector. There is no net force on the test charge.

What direction will a positive charge move, when introduced into a uniform electric field pointing directly upwards?

Directly upwards.

A positive charge always follows the electric field.

What direction will a negative charge move, when introduced into a uniform electric field pointing directly upwards?

Directly downwards.

A negative charge always goes exactly opposite to the field lines.

What direction will a non-charged particle move, when introduced into a uniform electric field pointing directly upwards?

No motion. A non-charged particle (neutral) will not be accelerated in any direction by an electric field.

What direction will a proton move, when introduced into a uniform electric field pointing towards the right?

To the right.

Positive charges always move in the direction of field lines.

What direction will an electron move, when introduced into a uniform electric field pointing towards the left?

To the right.

Negative charges move in the exact opposite direction of field lines.

**electric potential**

**Electric potential** (or **electric potential energy**) is the energy of position for charges, relative to each other.

This is the same as the electrostatic force between charges times the distance between them. Electric potential may be positive or negative, depending on the sign of the charges.

What is the relationship between charges and distance that gives electric potential?

**U = Kq _{1}q_{2} / r**

Where:

K = Coulombic constant in Nm^{2}/C^{2}

q_{1} and q_{2} = charges in C

r = distance between the charges in m

If the distance between two positive charges doubles, what will happen to the electic potential between them?

Electric potential will halve.

From: U = Kq_{1}q_{2} / r

U is inversely proportional to r, hence doubling r will halve U.

**potential difference** between two points

**Potential difference**, also called **voltage**, is the potential energy per charge that is lost or gained when a test charge is moved between two positions.

By convention, the moving charge is assumed to be a positive unit test charge of 1 Coulomb.

What is the relationship between charge and distance that gives potential difference (voltage)?

**V = K*q / r**

Where:

K = Coulombic constant in Nm^{2}/C^{2}

q = static charge in C

r = distance between the charge and a positive unit test charge in m

What change in voltage is necessary to hold a unit charge in position, if the main charge is suddenly doubled?

Voltage must double.

From: V = K*q / r

Voltage and main charge are directly proportional, hence doubling one will double the other.

**absolute potential** of a point in space

The **absolute potential **of a point in space is the voltage necessary to bring a test charge from infinity to that point, a given distance away from a fixed charge.

By convention, the test charge is assumed to be a positive unit charge of 1 Coulomb. This value can be positive or negative depending on the main charge.

Given the absolute potential for a fixed positive charge at a known position, what would be the absolute potential for a negative charge of the same magnitude at the same position?

The negative of the absolute potential value that was given for the positive charge.

Absolute potential will be positive or negative depending on the sign of the charge.

What concept describes the set of positions of equal energy that a positive test charge can be placed, when put next to a stationary charge?

**Equipotential lines. **

All charges have equipotential lines that form closed, non-overlapping loops around them.

A solitary 5C charge is fixed in space and a test charge is placed at 1cm and then 2cm away. Can these two positions be on the same equipotential line?

No. These are not positions where the potential electrostatic energy is constant.

From: energy = Kq_{1}q_{2} / r , the potential energy is inversely proportional to the distance between the fixed charge and the test charge. Doubling the distance will halve the energy, hence these positions cannot be equal energy and cannot be equipotential.

an **electric dipole**

An** electric dipole** occurs as the result of any distinct separation of positive and negative charge in an object or system.

On the AP Physics exam, the simplest case is always assumed: a linear object with one positive and one negative end.

Describe the motion of a dipole placed into an electric field.

The dipole will rotate so that the positive end is aligned with the electric field and the negative end is exactly opposite to the field.

Assuming that the dipole is overall neutral, there will not be any translation in the field.

**electric current, I**

**Electric current, I,** is the directional flow of charge through a conducting medium.

Though the moving charge in a circuit is typically electrons, the conventional direction of current always follows positive charge flow.

What is the formula and SI unit for current?

Current: * I* = ∆Q / ∆t

Where:

∆Q = change in charge (coulombs)

∆t = change in time (seconds)

The SI unit of charge is the **ampere, A,** and represents 1 C/s.

If electrons are moving along a wire from left to right, where is conventional current moving?

Conventional current is moving from right to left.

Conventional current always refers to the flow of positive charge, and will always be opposite to the flow of electrons.

By what proportion will the current change if the amount of charge transferred doubles and the time halves?

The current will be 4x larger.

*I _{0} *= Q / t

given new Q' = 2Q

and new t' = (1/2)t

then:

*I'*= Q' / t'

= 2Q / (1/2)t = 4(Q / t)

= 4

*I*

_{0}
a **battery**

An **electrical battery** (or electrochemical cell) is a device that produces a flow of electrons from anode to cathode.

The necessary electrical energy is created by undergoing redox chemical reactions in the cell.

**electromotive force (EMF)**

**Electromotive force** is the ability to transfer a unit of charge over the electrical potential difference between two electrodes.

This is not actually a "force". EMF is an energy per charge, i.e. work done to move charge (Volts).

**electric potential energy**

**Electric potential energy (or electrostatic potential energy) **is the energy, either for repulsion or attraction, between specific charges.

Technically, electric potential can be calculated between electric fields as well, but this is beyond the scope of the AP Physics exam.

What is the formula and SI unit of electric potential energy?

U = kQq/r

Where:

k = Coulomb's constant (9x10^{9} Nm^{2}/C^{2})

Q = primary charge in C

q = test (secondary) charge in C

r = radius between charges in m

As with all forms of energy, the SI unit is the **Joule, J,** and represents 1 N*m.

What does a positive vs. a negative electric potential energy indicate about the charges and force between them?

A **positive energy **happens when both charges are the same parity (+,+ or –,–), and will cause a repulsive force between the two charges.

A** negative energy** happens when both charges are opposite parity, and will cause an attractive force between the two charges.

**voltage**

**Voltage (electric potential)** is the electrical energy per unit charge necessary to move a test charge against the electric field of a fixed charge.

What is the formula and SI unit of voltage?

V = kQ/r

Where:

k = Coulomb's constant (9x10^{9} Nm^{2}/C^{2})

Q = primary charge in C

r = radius between charges in m

Voltage is in SI units of **volts, V, **and represents 1 J/C.

What is the change in voltage if a test charge at a distance r from a central charge Q is now moved twice as far away?

The voltage will decrease by 1/2.

From: V_{0}=kQ/r

new r'=2r

k and Q are constant

V''=kQ/r'=kQ/2r

=(kQ/r)(1/2)=V_{0}/2

**resistance**

**Resistance **is a measure of how difficult it is to pass current through some object. Many factors affect resistance, including: cross-sectional area, length, and general resistivity.

Though all materials have some resistance at room temperature, super-cooled superconductors have a resistance of zero.

What is the formula for calculating the total resistance of an object, and what are the SI units of resistance?

R = *ρ****L / A

Where:

*ρ* = resistivity of the material in Ωm

L = length of the object along current in m

A = cross sectional area in m^{2}

Resistance is in SI units of **Ohms, Ω**, and represents 1 J*s/C^{2}.

By what factor has total resistance changed if the length of a resistor is doubled?

The total resistance will also double.

From: R_{0} = ρ*L/A

new L' = 2L

R' = ρ*L'/A

= ρ*2L/A = 2(ρ*L/A)=2R_{0}

By what factor has total resistance changed if the cross-sectional area of a resistor is doubled?

The total resistance will decrease by 1/2.

From: R_{0} = ρ*L/A

new A' = 2A

R' = ρ*L/A'

= ρ*L/2A = (ρ*L/A)(1/2)=R_{0}/2

By what factor has total resistance changed if the length of a resistor doubles and the cross sectional area also doubles?

The total resistance will remain constant.

From: R_{0} = *ρ**L/A

new L' = 2L

new A' = 2A

R' = *ρ**L'/A'

= *ρ**2L/2A = *ρ**L/A=R_{0}

By what factor has total resistance changed of a wire, if the length doubles and the radius also doubles?

The total resistance will decrease by 1/2.

From: R_{0} = *ρ**L/A

new L' = 2L

new A' = 4A (since area goes as the square of the radius)

R' = *ρ**L'/A'

= *ρ**2L/4A = *ρ**L/2A=R_{0}/2

**Internal resistance** of a battery

**Internal resistance** of a battery is the difference between the predicted ideal voltage that a battery should deliver vs. the real-world value that it provides.

Occasionally, internal resistance may be referred to as the "impedence", since factors like battery size, chemical makeup, and current load can all *impede *ideal voltage.

**Resistivity**

**Resistivity, ρ,** is the measurement of how much a certain material resists the flow of current, and has SI units of **Ω*m.**

What is the formula for** Ohm's law?**

**Ohm's law **is written as:

*I* = V/R

or commonly: V = *I**R

Where:

V = voltage in V

*I* = current in A

R = resistance in Ω

What must happen to the resistance, if the voltage is doubled in a circuit that has constant current flow?

The resistance must double also.

From Ohm's law: V=*I*R

Since current is held constant, voltage and resistance are directly proportional. Doubling voltage must result in resistance doubling.

What must have happened to the voltage supplied to a fixed resistance circuit, if the current suddenly drops to half original?

The voltage must have also dropped to half.

From Ohm's law: V=*I*R

Since R is held constant, V and *I* are directly proportional. Halving voltage means current must be halved also.

What must have happened to the resistance in a fixed voltage circuit, if the current suddenly doubles?

The resistance must have dropped to 1/2 original.

From Ohm's law: V=*I*R

Since V is held constant, R and *I* are inversely proportional. Doubling current means resistance must be halved.

What does this symbol stand for in a circuit diagram?

This is a battery. Batteries will always have a larger positive terminal (labeled +) and a smaller negative terminal (labeled –).

Conventional current always flows from + to – around the circuit.

What does this symbol stand for in a circuit diagram?

This is a resistor. Resistors impede current flow and have units of ohms.

What does this symbol stand for in a circuit diagram?

This is a capacitor. Capacitors store charge and have units of farads.

What does this symbol stand for in a circuit diagram?

This is a switch. Switches can be open (as shown) so that no current flows around the circuit, or closed so that both dots are connected and current flows.

What does this symbol stand for in a circuit diagram?

This is the symbol for ground. Ground is the physical end of the circuit path and represents zero current or zero voltage.

If no ground symbol is given, the negative terminal of the battery is assumed to be ground.

What is the difference *visually *between resistors in parallel vs. series in a circuit?

Resistors in parallel are each on a diverging circuit pathway. A single charge will pass through either one or the other to complete the circuit. (first image)

Resistors in series follow each other on continuous circuit pathway. A single charge must go through both to complete the circuit. (second image)

What is the formula for calculating the total resistance of a circuit, for resistors in parallel?

1/R_{Total} = 1/R_{1} + 1/R_{2} + ...

Where:

R_{Total} = total resistance in Ω

R_{1} = first resistor in Ω

R_{2} = second resistor in Ω

... = any additional resistors in similar fashion

What is the total resistance in the circuit below?

R_{Total} = 5 Ω

1/R_{Total }= 1/R_{1} + 1/R_{2
}= 1/10 + 1/10 = 2/10

= 1/5

What is the formula for calculating the total resistance of a circuit, for resistors in series?

R_{Total} = R_{1} + R_{2} + ...

Where:

R_{Total} = total resistance in Ω

R_{1} = first resistor in Ω

R_{2} = second resistor in Ω

... = any additional resistors in similar fashion

What is the total resistance in the circuit below?

R_{Total} = 10 Ω

R_{Total} = R_{1} + R_{2 }

= 4 + 6 = 10

What is the total resistance in the circuit below?

The total resistance is 5Ω.

First find resistance for the 4Ω and 6Ω in series.

R_{series} = 4 + 6 = 10Ω.

Then find resistance for the 10Ω and 10Ω in parallel.

1/R_{Total} = 1/10 + 1/10

= 2/10 = 1/5.

R_{Total} = 5Ω.

What amount of current must be flowing around the circuit below?

The current is 2A.

First find resistance for the 4Ω and 6Ω in series.

R_{series} = 4 + 6 = 10Ω.

Then find resistance for the 10Ω and 10Ω in parallel.

1/R_{Total} = 1/10 + 1/10

= 2/10 = 1/5.

R_{Total }= 5Ω.

Finally, from Ohm's law:

*I* = V/R = 10/5 = 2A

What amount of current must be flowing around the circuit below?

The current is 1A.

First find resistance for the 12Ω and 6Ω in parallel.

1/R_{Total }= 1/12 + 1/6

= 3/12 = 1/4.

R_{Total} = 4Ω.

Then, from Ohm's law:

*I* = V/R = 4/4 = 1A.

**Capacitance**

**Capacitance** is the measured ability of a material to store charge.

Generally capacitance implies "parallel plate capacitance", where charge is related to the area of the plates, distance between plates, and delectric material between plates.

What is the formula and SI unit for capacitance?

Capacitance: C = Q / V

Where:

C = capacitance

Q = charge transferred

V = voltage

Capacitance is in SI units of **farads, F**, and represents 1 C/V.

What has happened to the capacitance, if the charge on a capacitor plate doubles while the voltage remains constant?

Capacitance will also double.

From capacitance: C_{0} = Q / V

new Q' = 2Q

C' = Q'/V = 2Q/V = 2(Q/V)= 2C_{0}

What is the formula for calculating the total capacitance of a circuit, for capacitors in parallel?

C_{Total} = C_{1} + C_{2} + ...

Where:

C_{Total }= total capacitance in F

C_{1} = first capacitor in F

C_{2} = second capacitor in F

... = any additional capacitors in similar fashion

What is the total capacitance in the circuit below?

Capacitance is 10F.

C_{Total} = C_{1} + C_{2}

= 8 + 2 = 10F

What is the formula for calculating the total capacitance of a circuit, for capacitors in series?

1/C_{Total} = 1/C_{1} + 1/C_{2} + ...

Where:

C_{Total} = total capacitance in F

C_{1} = first capacitor in F

C_{2} = second capacitor in F

... = any additional capacitors in similar fashion

What is the total capacitance in the circuit below?

Capacitance is 2F.

1/C_{Total} = 1/C_{1} + 1/C_{2}

= 1/4 + 1/4 = 2/4 = 1/2.

C_{Total} = 2F.

a **dielectric **material

A **dielectric material **is an electrical insulator, and decreases the electric field produced by a given charge.

All materials have a dielectric constant, k, which defines how effective that material is in decreasing the electric field.

Water has a dielectric constant k = 80.4 and glycerine has a dielectric constant of k = 42.5. Which of these two substances would lessen a charge's electric field more?

Water is a better dielectric than glycerin.

Recall: a higher k value indicates that substance is more effective at insulating a charge and lessening a charge's field.

What is the formula for capacitance between parallel plates?

C=k*ε_{0}*A / d

Where:

C = capacitance

A = area of overlap of the two plates

k = dielectric constant of the material between the plates (for a vacuum, k = 1)

ε_{0} = electric constant (ε_{0} = 8.85×10^{−12} F/m)

d = distance between the plates

What will happen to capacitance, if a dielectric with k>1 is inserted between the parallel plates of a capacitor?

Capacitance increases proportional to k.

Since: C=k*ε_{0}*A / d

and for air k=~1

any k value greater than 1 will directly increase capacitance.

What has happened to the voltage of a fixed circuit, if a dielectric material is inserted between plates of a capacitor and the charge on those plates doubles in order for current to continue to flow?

Voltage provided by a battery is always constant in a fixed circuit.

Capacitance of the circuit will have increased; but since current is still flowing, it must be true that nothing has been done to change the voltage of the battery.

What is the formula for calculating the energy stored in a charged capacitor?

E = (1/2)CV^{2}

Where:

C = capacitance in farads

V = voltage of the circuit in volts

Note: energy stored in a capacitor is necessarily the same as the work done initially to charge it.

What is the energy necessary to charge a 10F capacitor on a 6V circuit?

180 J

From E = (1/2)CV^{2}

C = 10F, V = 6V.

E = (1/2)*10*6^{2}

= (1/2)*10*36=180J.

What has happened to the capacitance of a fixed voltage circuit, if a dielectric material is inserted between plates of a capacitor and the charge on those plates doubles in order for current to continue to flow?

Capacitance will double.

From the formula: C = Q/V

Since V is held constant, Q and C are directly proportional. Doubling charge must mean that capacitance doubles also.

**Conductivity**

**Conductivity** is the measure of how well a material is able to conduct electric current through it, and is represented by σ (sigma) or κ (kappa).

Conductivity is the inverse of resistivity and will have units of S/m, where S is the SI unit of siemens.

What must the conductivity of iron be, if it is found that the resistivity of iron is 10^{-7 }Ω*m?

10^{7} S/m

Since conductivity is the reciprocal of resistivity, conductivity = 1/10^{-7} = 10^{7}.

What must have happened to the conductivity of a metal, if raising the temperature halves the resistance?

The conductivity increases to double.

Conductance and resistance are inversely proportional, so halving one will double the other.

**Power**

**Power** is the rate at which energy is transferred per unit time and has SI units of **Watts (W),** which represents J/s.

What are the formulas for power transferred through a circuit and power dissipated by a circuit?

Power transferred is P = IV

Power dissipated is P = I^{2}R

Where:

I = current in A

V = voltage in V

R = resistance in Ω

How will the amount of power required to transfer charges change, if the voltage of the circuit doubles while the current halves?

Power required is the same.

From P_{0} = IV

V' = 2V, I' = I/2

P' = I'V' = (I/2)(2V) = IV = P_{0}

How will the amount of power dissipated by a circuit change, if the resistance remains constant while the current doubles?

Power dissipated will be 4x original.

From P = I^{2}R

With resistance constant, doubling current will quadruple power, since power is directly proportional to the square of current.

**RMS**

**RMS** stands for **Root Mean Square**, and represents the relationship between the instantaneous peak value and the average (mean) value.

RMS can be calculated by taking the peak value and dividing by √2.

In the image below, the peak current is 1A, so the RMS current will be 1/√2 = .7A

**AC current**

AC current (alternating current) periodically reverses direction of flow, creating both positive and negative values.

This is different from DC (direct current), which flows in only one direction and remains always positive.

Note: to make an AC current value into a DC current equivalent, divide the peak AC current by √2, this is known as the RMS factor.

**AC Voltage**

**AC voltage** (alternating voltage) periodically reverses sign, having both positive and negative values, due to AC current changing direction and sign.

This is different from normal voltage (direct voltage), which remains always positive, due to supposing that current flows in one direction.

Note: to make an AC voltage value into a DC voltage equivalent, divide the peak AC voltage by √2, this is known as the RMS factor.

**AC Power**

**AC power** (alternating power) is the energy transferred by an alternating current.

This is similar to normal power, the only difference is that all equation variables are the RMS equivalents.

P_{RMS} = I_{RMS}*V_{RMS}

P_{RMS} = I^{2}_{RMS}*R

How will the amount of power required to transfer charges change, if the voltage of an AC circuit halves while the current doubles?

Power will remain the same as original.

From P_{RMS} = V_{RMS}*I_{RMS}

V' = V/2, I' = 2I

P' = V'I' = (V/2)*2I = VI = P_{RMS}

an **electrical circuit switch**

An **electrical switch **is a circuit component that can break current pathways, or change between pathways.

What is the amount of current flowing through the 10Ω resistor is the circuit as currently shown below?

Current is 1A through the 10Ω resistor.

Since the switch is open in the circuit, no current flows through either the 4Ω or the 6Ω resistor. This means that the 10Ω resistor is the only one allowing current around the 10V circuit.

I = V/R = 10/10 = 1A

What voltage battery is necessary in order for 4A of current to be flowing through the circuit below?

Voltage must be 24V.

First, find total resistance:

1/R_{Total} = 1/R_{1} + 1/R_{2}

= 1/8 + 1/24 = 4/24

= 1/6.

R_{Total }= 6Ω

Then, From Ohm's law:

V=*I*R = 4*6 = 24V.

In the circuit below, the total amount of current flowing is 1A. If both the 8Ω and 6Ω resistors were removed and the same battery used, by what factor will the current be changed?

The new current will be 8x increased.

Original resistance was

R = 8 + 1/(1/4+1/4) + 6

= 8 + 2 + 6 = 16Ω.

I = V/R = V/16.

New resistance is 1/(1/4+1/4) = 2Ω.

I' = V/R' = V/2.

Comparing I : I' = 1:8.