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AP Physics C Electricity : Electricity and Magnetism Exam

Study concepts, example questions & explanations for AP Physics C Electricity

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Example Questions

Example Question #2 : Electromagnetism

A wire loop of area is placed in a magnetic field whose magnitude varies with time according to the equation $B(t) = B_o sin(\omega t)$ . The loop is held in place, perpendicular to the field. If the wire has resistance , what is the magnitude of the maximum instantaneous current induced in the loop?

Possible Answers:

$\frac{B_o A}{\omega R}$

$\frac{B_o\omega A}{R}$

$\frac{B_o sin(\omega) A}{R}$

$\frac{B_o \omega}{AR}$

$\frac{B_o cos(\omega ) A}{R}$

Correct answer:

$\frac{B_o\omega A}{R}$

Explanation:

Relevant equations:

$|\epsilon| = \frac{d\phi}{dt} = \frac{d (B\cdot A)}{dt}$

$\phi = B \cdot A$

$\epsilon = IR$

Write the expression for the flux using the given expression for the magnetic field strength:

$\phi = B\cdot A = B_o sin(\omega t)A$

Take the derivative of this flux to find the induced emf:

$|\epsilon | = \frac{d\phi}{dt} = \frac{d (B_o sin(\omega t) A)}{dt} = B_o \omega cos(\omega t) A$

Determine the maximum value of $cos(\omega t)$ , and replace $cos(\omega t)$ with this value, to find the maximum induced emf:

$|\epsilon | _{max} = {B_o \omega * 1 * A} = B_o \omega A$

Plug this maximum induced emf into $\epsilon = IR$ , to find the maximum induced current:

$\epsilon = IR \rightarrow I = \frac{\epsilon}{R} = \frac{B_o \omega A}{R}$

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Example Question #201 : Ap Physics C

A wire loop carrying no current is placed near another very long straight wire carrying current . The loop is then moved directly away from the straight wire at a constant speed.

Ps0_lenz

In which direction will the current induced in the small wire loop flow?

Possible Answers:

Counterclockwise

No current will be induced in the wire loop

Clockwise

Not enough information is given to solve

Correct answer:

Clockwise

Explanation:

The magnetic field due to the long straight wire points into the page (right-hand rule) in the region below it in the picture. The magnitude of this magnetic field is given by:

$B=\frac{\mu _{oI}}{2\pi r}$

The magnetic field thus decreases in value with increasing distance, . As such, the amount of flux passing through the wire loop decreases steadily as it is moved away from the wire.

According to Lenz's Law, an induced current within the wire loop will counteract this loss of magnetic flux from the straight wire. The magnetic field from the induced current in the wire loop will consequently point in the same direction as that from the current in the straight wire. The right-hand rule for a magnetic field into the page yields an induced current direction in the clockwise direction.

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Example Question #4 : Electromagnetism

A capacitor and an inductor mH are connected in series to complete a simple circuit. The capacitor is initially charged to 9.9 nC.

Ps0_lccircuit

$C=820nF,\ L=530mH$

What is the value of the maximum current that passes through the circuit?

Possible Answers:

$8.4*10^{-7}A$

3.0A

$6.1 * 10^{-4}A$

$1.5 * 10^{-5 }A$

Correct answer:

$1.5 * 10^{-5 }A$

Explanation:

Since there are no dissipative circuit elements (such as a resistor) present in the circuit, the energy originally held by the electric field of the capacitor oscillates between electric energy and magnetic energy over time. As such, the maximum electric energy of the capacitor is equal in value to the maximum magnetic energy of the inductor.

$E_{C_{max}}=\frac{1}{2}\frac{(Q_{max})^2}{C}$

$E_{L_{max}}=\frac{1}{2}L(I_{max})^2$

$\frac{1}{2}\frac{(Q_{max})^{2}}{C}=\frac{1}{2}L(I_{max})^{2}$

$\frac{(9.9nC)^2}{820nF}=(530mH)(I_{max})^2$

$I_{max}=1.5*10^{-5}A$

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Example Question #1 : Electromagnetism

A uniform magnetic field is created in a region in space, as represented by the grey arrows in the diagram below. Four circular loops A, B, C, and D are placed in the magnetic field. The planes of loops A and B are oriented perpendicularly to the direction of the field; the planes of loops C and D are oriented parallel to the field.

Ps0_fluxcomparison

Loops A and C are the same size, with radius . Loops B and D are the same size, with radius . Note that R>r .

Which of the statements below is true about the values of magnetic flux through the loops due to the magnetic field?

Possible Answers:

$\phi_{A}=\phi_{C}<\phi_{B}=\phi_{D}$

$\phi_{C}=\phi_{D}<\phi_{A}<\phi_{B}$

$\phi_{A}<\phi_{B}<\phi_{C}<\phi_{D}$

$\phi_{C}<\phi_{A}<\phi_{D}<\phi_{B}$

Correct answer:

$\phi_{C}=\phi_{D}<\phi_{A}<\phi_{B}$

Explanation:

Magnetic flux is a measure of the amount of magnetic field passing through a given area and is given by the equation:

$\phi_{m} =\int \overrightarrow{B}\cdot \overrightarrow{dA}$

The magnetic field is constant for this problem, and the areas are all circular loops. As such, the vector product integral simplifies to:

$\phi_{m}=BA\;cos\, \theta$

The angle theta is the angle between a vector normal to the surface of the area and the given magnetic field.

For loops C and D, theta is 90 degrees, so the magnetic flux for both of these loops is equal to zero.

$\phi_{m}=BA\;cos(90^\circ)=BA(0)=0$

For loops A and B, the angle theta is 0 degrees, so the magnetic flux for both of these loops is nonzero and proportional to the area of each loop. Therefore, the flux for loop B is larger than that for loop A because loop B has the larger radius.

R>r

A_B>A_A

$\phi_{m_B}>\phi_{m_A}$

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Example Question #1 : Electric Circuits

A parallel plate capacitor has a capacitance of . If the plates are 1cm apart, what is the area of the plates?

$\epsilon_0=8.85* 10^{-12} \frac{F}{m}$

Possible Answers:

1.1*10^7m^2

$1.1*10^{11}m^2$

$1.1*10^{10}m^2$

1.1*10^9m^2

Correct answer:

1.1*10^9m^2

Explanation:

The relationship between capacitance, distance, and area is $C=\epsilon_0 \frac{A}{d}$ . We can rearrange this equation to solve for area.

$A=\frac{Cd}{\epsilon_0}$

Now, we can use the values given in teh question to solve.

$A=\frac{1F(1*10^{-2}m)}{8.85* 10^{-12} \frac{F}{m}}=1.1*10^9m^2$

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Example Question #1 : Capacitors

Charge is distributed uniformly over the area of the two plates of a parallel plate capacitor, resulting in a surface area charge density of $\pm \sigma$ on the plates (the top plate is positive and the bottom is negative, as shown below). Each plate has area and are separated by distance . A material of dielectric constant $\kappa_{material}$ has been placed between the two plates.

Ps0_capacitor

Which of the following would not result in an increase in the measure of electric potential difference between the two plates?

Possible Answers:

Increase the value of the surface charge density, $\sigma$ , on each plate

Increase the area, , of the plates

Increase the distance, , between the plates

Replace the material between the two plates with one of a lower dielectric constant, $\kappa$

Correct answer:

Increase the area, , of the plates

Explanation:

The electric potential difference created between the plates of a parallel plate capactor is given by the equation:

$V = \frac{Q}{C}$

The charge can be calculated by using the equation:

$Q=\sigma A$

The value of the capacitance is related to the dimensions of the capacitor with the equation:

$C=\kappa \frac{\varepsilon _{o}A}{d}$

Combining these equations yields:

$V=\frac{\sigma A}{\frac{\kappa \epsilon_0A}{d}}=\frac{\sigma d}{\varepsilon_{o} \kappa }$

The area becomes inconsequential, while the potential is directly proportional to the surface charge density and the distance between the plates, and inversely proportional to the dielectric of the material between the plates. Changing the area does not cause any change in the potential difference measured between the plates, and changing any of the other variables would cause a resultant change in the potential difference.

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Example Question #203 : Ap Physics C

You are hired to make a capacitor out of two parallel metal sheets. If someone wanted you to make a thin capacitor of $700\mu F$ out of those metal sheets, and the sheets needed to be 0.5mm apart, what area do the two metal plates need to be?

$\epsilon_0=8.85*10^{-12}\frac{C^2}{Nm^2}$

Possible Answers:

6.21*10^6m^2

$3.95*10^{-4}m^2$

2.77*10^3m^2

5.47*10^8m^2

3.95*10^4m^2

Correct answer:

3.95*10^4m^2

Explanation:

For parallel plate capacitors, the equation is $C=\frac{\epsilon_0A}{d}$ .

Solve for .

$A=\frac{dC}{\epsilon_0}$

Now we can plug in our given values.

$A=\frac{(0.5*10^{-3})(700*10^{-6})}{8.85*10^{-12}\frac{\text{C}^2}{\text{Nm}^2}}$

$A=3.95*10^4\ \text{m}^2$

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Example Question #1 : Capacitors

The plates of a parallel plate capacitor are 3mm apart and 5m^2 in area. A potential difference of 1000V is applied across the capacitor. Find the capacitance.

$\epsilon_0=8.85*10^{-12} \frac{F}{m}$

Possible Answers:

$1.5*10^{-8}F$

$1.5*10^{-10}F$

$1.5*10^{-7}F$

$1.5*10^{-9}F$

Correct answer:

$1.5*10^{-8}F$

Explanation:

Capacitance is related to plate area and distance by the equation $C=\epsilon_0\frac{A}{d}$ .

Given the area and distance, we can solve for capacitance. The voltage, in this case, is irrelevant.

$C=8.85* 10^{-12} \frac{F}{m}(\frac{5 m^2}{3*10^{-3}m})=8.85* 10^{-12} \frac{F}{m}(1.67*10^{3}m)=1.5*10^{-8}F$

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Example Question #2 : Capacitors

The plates of a parallel plate capacitor are 3mm apart and 5m^2 in area. A potential difference of 1000V is applied across the capacitor. Compute the charge on each plate.

$\epsilon_0=8.85* 10^{-12} \frac{F}{m}$

Possible Answers:

$1.5*10^{-5}C$

$1.5*10^{-6}C$

$1.5*10^{-8}C$

$1.5*10^{-7}C$

Correct answer:

$1.5*10^{-5}C$

Explanation:

Charge on a capacitor is given by the equation Q=CV . We know that the voltage is 1000V , but we need to determine the capacitance based off of the area and distance between the plates.

$C=\epsilon_0\frac{A}{d}$

$C=(8.85*10^{-12}\frac{F}{m})(\frac{5m^2}{3*10^{-3}m})$

$C=(8.85*10^{-12}\frac{F}{m})(1.67*10^{3}m)=1.5*10^{-8}F$

We can plug this value into the equation for charge.

$Q=CV=(1.5*10^{-8}F)(1000V)$

$Q=1.5*10^{-5}C$

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Example Question #1 : Electric Circuits

The plates of a parallel plate capacitor are 3mm apart. A potential difference of 1000V is applied across the capacitor. Compute the magnitude of the generated electric field.

$\epsilon_0=8.85*10^{-12}\frac{F}{m}$

Possible Answers:

$3.33*10^6\frac{N}{C}$

$3.33*10^5\frac{N}{C}$

$3.33*10^4\frac{N}{C}$

$3.33*10^3\frac{N}{C}$

Correct answer:

$3.33*10^5\frac{N}{C}$

Explanation:

The electric field given by a capacitor is given by the formula $E=\frac{Q}{\epsilon_0 A}$ . We do not have these variables, so we will have to adjust the equation.

$Q=CV\rightarrow E=\frac{CV}{\epsilon_0A}$

The capacitance can be determined by the area of the plates and the distance between them.

$C=\epsilon_0\frac{A}{d}\rightarrow E=\frac{\epsilon_0\frac{A}{d}V}{\epsilon_0A}$

This equation simplifies to $E=\frac{V}{d}$ , allowing us to solve using the values given in the question.

$E=\frac{1000V}{3*10^{-3}m}=3.33*10^{5}\frac{N}{C}$

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All AP Physics C Electricity Resources

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AP Physics C Electricity : Electricity and Magnetism Exam

Study concepts, example questions & explanations for AP Physics C Electricity

All AP Physics C Electricity Resources

Example Questions

Example Question #2 : Electromagnetism

Example Question #201 : Ap Physics C

Example Question #4 : Electromagnetism

Example Question #1 : Electromagnetism

Example Question #1 : Electric Circuits

Example Question #1 : Capacitors

Example Question #203 : Ap Physics C

Example Question #1 : Capacitors

Example Question #2 : Capacitors

Example Question #1 : Electric Circuits

All AP Physics C Electricity Resources

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