All AP Physics B Resources
Example Questions
Example Question #2 : Ap Physics 2
Which of the following is not an acceptable unit for specific heat?
Specific heat is most commonly applied in the equation to determine the heat required/released during a temperature change:
Rearranging this equal, we can see that this units of specific heat are units of heat per units of mass times units of temperature.
Heat energy can have the units of Joules or calories. Mass can have units of grams or kilograms. Temperature can have units of Kelvin, degrees Celsius, or degrees Fahrenheit. British thermal units (Btu) is a less common unit of heat, defined by the amount of heat required to raise one pound of water by one degree Fahrenheit.
Newtons are a unit of force, and cannot be used in the units for specific heat.
Example Question #1 : Understanding Heat And Temperature
A sample of copper is heated with of heat and the temperature of the sample rises from to . What mass of copper is present in this sample?
Our equation for temperature change using specific heat is:
Since Celsius and Kelvin use the same interval scale, we do not need to convert to find the change in temperature. Using the initial and final temperatures, the energy input, and the specific heat of copper, we can calculate the mass of copper present.
To simplify, we will solve for the temperature change separately.
Since the scale for Celsius and Kelvin is the same, this change is the same in either scale.
Use this value back in the original calculation.
Convert to kilograms to simplify.
Example Question #1 : Thermodynamics And Heat
of heat energy is applied to a block of ice initially at . This is enough to change the ice completely to water. What is the final temperature of the water after the of heat is applied?
Since the ice melts, this action occurs in three stages.
1) The heating of the ice to its melting point
2) The melting of the ice into water
3) The heating of the water after it is melted
When a temperature change is involved, the relationship of heat input to temperature change is given by:
In this formula, is the specific heat (capacity) for the substance. For the phase change involved in melting, there is no temperature change (the energy goes into breaking bonds), and the amount of heat required to completely melt the substance is:
So this situation is encompassed by three expressions, representing the three parts of the problem, each of which deals with the same mass of material, . Keep in mind that the melting point for ice is , or .
1)
2) (for the ice completely melting)
3) (for the water warming up to its final temperature)
The values for , , and are given, and it is known that .
Using this equation we can solve for the final temperature.
Example Question #1 : Electricity And Magnetism
Charges A and B are placed a distance of from one another. The charge of particle A is whereas the charge of particle B is . Charge B experiences an electrostatic force of from charge A. Similarly, charge A experiences an electrostatic force of from charge B.
A student places these two charges in a vacuum medium and measures . A second student places these two charges in a glass medium and measures . What is ratio of the force measured in the glass medium to the force measured in the vacuum?
The dielectric constant of glass is .
You can simplify this question tremendously by using the definition of a dielectric constant. Dielectric constant is defined as the ratio of the electrostatic force in vacuum to the electrostatic force in the medium (in this case glass).
The question is asking for the reciprocal of this value: the ratio of the force in glass to the force in the vacuum. Our answer is calculated by taking the reciprocal of the dielectric constant of glass.
Example Question #1 : Electricity And Magnetism
How much work is done by the electric field moving an electron along an equipotential surface with a potential of ?
We must know the strength of the electric field to solve
Moving a charge along an equipotential surface will involve no work done by the electric field. The potential is constant throughout an equipotential surface; therefore, the potential difference experienced by the electron will be zero. Remember that energy is dependent on the potential difference.
If the potential difference is zero then the energy (and work) will be zero.
Example Question #2 : Coulomb's Law
Charges A and B are placed a distance of from one another. The charge of particle A is whereas the charge of particle B is . Charge B experiences an electrostatic force of from charge A. Similarly, charge A experiences an electrostatic force of from charge B.
What is the ratio of to ?
This question is very simple if you realize that the force experienced by both charges is equal.
The definition of the two electrostatic forces are given by Coulomb's law:
In this question, we can rewrite this equation in terms of our given system.
It doesn’t matter if the charges of the two particles are different; both particles experience the same force because the charges of both particles are accounted for in the electrostatic force equation (Coulomb's law). This conclusion can also be made by considering Newton's third law: the force of the first particle on the second will be equal and opposite the force of the second particle on the first.
Since the forces are equal, their ratio will be .
Example Question #12 : Electricity
An excess charge of is put on an ideal neutral conducting sphere with radius . What is the Coulomb force this excess charge exerts on a point charge of that is from the surface of the sphere?
Two principal realizations help with solving this problem, both derived from Gauss’ law for electricity:
1) The excess charge on an ideal conducting sphere is uniformly distributed over its surface
2) A uniform shell of charge acts, in terms of electric force, as if all the charge were contained in a point charge at the sphere’s center
With these realizations, an application of Coulomb’s law answers the question. If is the point charge outside the sphere, then the force on is:
In this equation, is Coulomb’s constant, is the excess charge on the spherical conductor, and is total distance in meters of from the center of the conducting sphere.
Using the given values in this equation, we can calculate the generated force:
Example Question #2 : Magnetic Force
A proton traveling at in a horizontal plane passes through an opening into a mass spectrometer with a uniform magnetic field directed upward. The particle then moves in a circular path through and crashes into the wall of the spectrometer adjacent to the entrance opening. How far down from the entrance is the proton when it crashes into the wall?
The proton’s mass is and its electric charge is .
A charged particle moving through a perpendicular magnetic field feels a Lorentz force equal to the formula:
is the charge, is the particle speed, and is the magnetic field strength. This force is always directed perpendicular to the particle’s direction of travel at that moment, and thus acts as a centripetal force. This force is also given by the equation:
We can set these two equations equal to one another, allowing us to solve for the radius of the arc.
Once the particle travels through a semicircle, it is laterally one diameter in distance from where is started (i.e. twice the radius of the circle).
Twice this value is the lateral offset of its crash point from the entrance:
Example Question #1 : Electromagnetism
An ideal transformer, under a purely resistive load, consists of an iron core that directs all the magnetic flux from a primary winding through a secondary winding. The primary winding has closely-packed loops along the coil (such that all see the same magnetic flux). The primary line carries an alternating current at a potential of . If the secondary line is to have a potential of , how many closely-packed loops are required in the secondary winding?
In basic consideration, the electrical transformer works directly according to Faraday’s law, following the picture given below.
If the coils are wound tightly enough that the magnetic flux generated in any of them effectively passes undiminished through all of them, and the iron core directs all the magnetic flux from the primary to secondary coil with insignificant losses, then Faraday’s law points a direct path from to . Faraday’s law relates the time of rate of change of magnetic flux to the voltage (emf) around a closed loop:
If there are loops stacked together, then the total voltage, , is related to the flux through all loops by:
From the figure, it is seen that the voltages in each line are dropped over all loops together. Hence, for the primary winding, by Faraday’s law:
For the secondary winding:
Here, and are the number of coils in the primary and secondary windings. The quantity is the same in both windings, being transmitted fully (and effectively instantly) by the iron core. This leads to the equation:
This can be used to solve the problem above:
NOTE: This only works under alternating current. Otherwise, there is no change in current to create the changing magnetic flux on which the transformer depends.
Example Question #71 : Ap Physics B
Which of the following is equal to the time constant of an RC circuit?
The definition of a time constant for an RC circuit is the product of resistance and charge:
This is defined as the time it takes the capacitor to reach a charge that is around 63% of the maximum charge. It is also the time it takes to discharge around 37% of the charge. This value is commonly used to determine the amount of charge a capacitor has or the amount of current flowing through the capacitor at any given time point.