# AP Physics B : Electricity and Magnetism

## Example Questions

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### Example Question #5 : Ap Physics 2

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 .

Explanation:

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 #6 : Ap Physics 2

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

Explanation:

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 #7 : Ap Physics 2

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 ?

Explanation:

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 #14 : Electrostatics

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?

Explanation:

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 #3 : 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 .

Explanation:

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 : Electricity And Magnetism

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?

Explanation:

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 #1 : Understanding Capacitors

Which of the following is equal to the time constant of an RC circuit?

Explanation:

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.

### Example Question #9 : Ap Physics 2

Which of the following statements is false regarding capacitors?

I. A larger capacitor stores more energy

II. Capacitors connected in parallel have the same voltage drop

III. Increasing the distance between the two plates increases the capacitance

II and III

I and III

III only

I only

I and III

Explanation:

Recall the equation for potential energy in terms of capacitance:

According to the equation above, an increase in capacitance will lead to decrease in the amount of energy stored. This means that a larger capacitor will store less energy; therefore, statement I is false.

Any circuit elements connected in parallel will always have the same voltage drop. This applies to resistors, capacitors, and inductors. Statement II is true.

Remember that capacitance is defined as follows:

In this equation, is the area of the plates and  is distance between the plates. Increasing  will lead to a decrease in capacitance; therefore, statement III is false.

### Example Question #10 : Ap Physics 2

A capacitor has a stored charge of  and a potential energy of . What is the voltage across the capacitor?

Explanation:

To solve this problem, you need to use the following two equations:

By substituting into the second equation you can write the change in potential energy in terms of voltage and charge.

Rearranging the previous equation will allow us to solve for the voltage.

Use the given values for stored potential energy and charge to solve for the voltage.

### Example Question #1 : Electric Circuits

battery is hooked to a circuit with a  and a  resistor in series with each other, and with a branched set of three parallel resistors of , , and . What current does this circuit draw from the battery?

Explanation:

Our first step will be to find the equivalent resistance of the circuit. Simple resistors add in series, and add inversely in parallel. For the  and  resistors in series, the total resistance is simply the sum.

For the resistors in parallel, the inverse of the total resistance is equal to the sum of the inverse of the individual resistors.

The total resistance of the circuit will be the sum of the series resistance and parallel resistance.

Now we know the total resistance of the circuit and the voltage supplied from the battery. Use Ohm's law to solve for the current in the circuit.

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