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AP Physics 2 : AP Physics 2

Study concepts, example questions & explanations for AP Physics 2

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All AP Physics 2 Resources

6 Diagnostic Tests 149 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Electric Fields

Coloumbs law for varsity

Charge has a charge of 9 nC

Charge has a charge of -8nC

The distance between their centers, is 3 mm .

What is the magnitude of the electric field at the center of due to

Possible Answers:

$|\overrightarrow{E}|=6.9*10^6\frac{N}{C}$

$|\overrightarrow{E}|=9.5*10^6\frac{N}{C}$

$|\overrightarrow{E}|=2.21*10^6\frac{N}{C}$

None of these

$|\overrightarrow{E}|=8*10^6\frac{N}{C}$

Correct answer:

$|\overrightarrow{E}|=8*10^6\frac{N}{C}$

Explanation:

Using the electric field equation:

$\overrightarrow{E}=k\frac{q}{r^2}$

Where is $9.0*10^9 N\frac{m^2}{C^2}$

is the charge, in coulombs

is the distance, in meters .

Convert to and plug in values:

$\overrightarrow{E}=9.0*10^9\frac{-8*10^{-9}}{.003^2}$

$\overrightarrow{E}=8*10^6\frac{N}{C}$

Magnitude is equivalent to absolute value:

$|\overrightarrow{E}|=8*10^6\frac{N}{C}$

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Example Question #18 : Electrostatics

Coloumbs law for varsity

Charge has a charge of 19 nC

Charge has a charge of -28nC

The distance between their centers, is 4.5 mm .

What is the magnitude of the electric field at the center of due to ?

Possible Answers:

$|\overrightarrow{E}|=8.44*10^6\frac{N}{C}$

$|\overrightarrow{E}|=3.98*10^6\frac{N}{C}$

None of these

$|\overrightarrow{E}|=6.11*10^6\frac{N}{C}$

$|\overrightarrow{E}|=3.75*10^6\frac{N}{C}$

Correct answer:

$|\overrightarrow{E}|=8.44*10^6\frac{N}{C}$

Explanation:

Using the electric field equation:

$\overrightarrow{E}=k\frac{q}{r^2}$

Where is $9.0*10^9 N\frac{m^2}{C^2}$

is the charge, in coulombs

is the distance, in meters .

Convert to and plug in values:

$\overrightarrow{E}=9.0*10^9\frac{19*10^{-9}}{.0045^2}$

$\overrightarrow{E}=8.44*10^6\frac{N}{C}$

Magnitude is equivalent to absolute value:

$|\overrightarrow{E}|=8.44*10^6\frac{N}{C}$

Report an Error

Example Question #1 : Electric Fields

Coloumbs law for varsity

Charge has a charge of 35 nC

Charge has a charge of -48nC

The distance between their centers, is 25 mm .

What is the magnitude of the electric field at the center of due to

Possible Answers:

$|\overrightarrow{E}|=2.25*10^5\frac{N}{C}$

$|\overrightarrow{E}|=3.77*10^5\frac{N}{C}$

$|\overrightarrow{E}|=1.73*10^5\frac{N}{C}$

None of these

$|\overrightarrow{E}|=6.91*10^5\frac{N}{C}$

Correct answer:

$|\overrightarrow{E}|=6.91*10^5\frac{N}{C}$

Explanation:

Use the electric field equation:

$\overrightarrow{E}=k\frac{q}{r^2}$

Where is $9.0*10^9 N\frac{m^2}{C^2}$

is the charge, in coulombs

is the distance, in meters .

Convert to and plug in values:

$\overrightarrow{E}=9.0*10^9\frac{-48*10^{-9}}{.025^2}$

$\overrightarrow{E}=-6.91*10^5\frac{N}{C}$

Magnitude is equivalent to absolute value:

$|\overrightarrow{E}|=6.91*10^5\frac{N}{C}$

Report an Error

Example Question #2 : Electric Fields

What is the electric field strength of a stationary 30C charge at a distance of 80cm away?

$k=8.9875*10^{9}\: \frac{N\cdot m^{2}}{C^{2}}$

Possible Answers:

$4.213*10^{11} \frac{N}{C}$ pointing towards the source charge

$3.370*10^{11} \frac{N}{C}$ pointing towards the source charge

$3.370*10^{11} \frac{N}{C}$ pointing away from the source charge

$4.213*10^{11} \frac{N}{C}$ pointing away from the source charge

Correct answer:

$4.213*10^{11} \frac{N}{C}$ pointing away from the source charge

Explanation:

To solve this question, we need to recall the equation for electric field strength.

$E=k\frac{q_{source}}{r^{2}}$

Notice that the equation above represents an inverse square relationship between the electric field and the distance between the source charge and the point of space that we are interested in.

Plug in the values given in the question stem to calculate the magnitude of the electric field.

$E=\left(8.9875*10^{9}\: \frac{N\cdot m^{2}}{C^{2}}\right)\frac{30 C}{(0.80 m)^{2}}=4.213*10^{11} \frac{N}{C}$

Now that we have determined the magnitude of the electric field, we need to identify which direction it is pointing with respect to the source charge. To do this, we'll need to remember that electric fields point away from positive charges and towards negative charges. Therefore, since our source charge is positive, the electric field will be pointing away from the source charge.

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Example Question #11 : Electric Fields

Coloumbs law for varsity

Charge has a charge of 16 nC

Charge has a charge of -25nC

The distance between their centers, is 7 mm .

What is the magnitude of the electric field at the center of due to

Possible Answers:

$|\overrightarrow{E}|=2.33*10^6\frac{N}{C}$

None of these

$|\overrightarrow{E}|=1.44*10^6\frac{N}{C}$

$|\overrightarrow{E}|=1.61*10^6\frac{N}{C}$

$|\overrightarrow{E}|=4.59*10^6\frac{N}{C}$

Correct answer:

$|\overrightarrow{E}|=4.59*10^6\frac{N}{C}$

Explanation:

Use the electric field equation:

$\overrightarrow{E}=k\frac{q}{r^2}$

Where is $9.0*10^9 N\frac{m^2}{C^2}$

is the charge, in Coulombs

is the distance, in meters.

Convert to and plug in values:

$\overrightarrow{E}=9.0*10^9\frac{-25*10^{-9}}{.007^2}$

$\overrightarrow{E}=-4.59*10^6\frac{N}{C}$

Magnitude is equivalent to absolute value:

$|\overrightarrow{E}|=4.59*10^6\frac{N}{C}$

Report an Error

Example Question #11 : Electric Fields

Coloumbs law for varsity

Charge has a charge of 9 nC

Charge has a charge of -8nC

The distance between their centers, is 3 mm .

What is the magnitude of the electric field at the center of due to ?

Possible Answers:

$|\overrightarrow{E}|=9*10^6\frac{N}{C}$

$|\overrightarrow{E}|=1.7*10^6\frac{N}{C}$

$|\overrightarrow{E}|=6.5*10^6\frac{N}{C}$

$|\overrightarrow{E}|=17*10^6\frac{N}{C}$

None of these

Correct answer:

$|\overrightarrow{E}|=9*10^6\frac{N}{C}$

Explanation:

Use the electric field equation:

$\overrightarrow{E}=k\frac{q}{r^2}$

Where is $9.0*10^9 N\frac{m^2}{C^2}$

is the charge, in Coulombs

is the distance, in meters.

Convert to and plug in values:

$\overrightarrow{E}=9.0*10^9\frac{9*10^{-9}}{.003^2}$

$\overrightarrow{E}=9*10^6\frac{N}{C}$

Magnitude is equivalent to absolute value:

$|\overrightarrow{E}|=9*10^6\frac{N}{C}$

Report an Error

Example Question #11 : Electric Fields

Coloumbs law for varsity

Charge has a charge of 25 nC

Charge has a charge of -38nC

The distance between their centers, is 15 mm .

What is the magnitude of the electric field at the center of due to

Possible Answers:

$|\overrightarrow{E}|=2.22*10^6\frac{N}{C}$

None of these

$|\overrightarrow{E}|=1.52*10^6\frac{N}{C}$

$|\overrightarrow{E}|=3.38*10^6\frac{N}{C}$

$|\overrightarrow{E}|=6.61*10^6\frac{N}{C}$

Correct answer:

$|\overrightarrow{E}|=1.52*10^6\frac{N}{C}$

Explanation:

Use the electric field equation:

$\overrightarrow{E}=k\frac{q}{r^2}$

Where is $9.0*10^9 N\frac{m^2}{C^2}$

is the charge, in Coulombs

is the distance, in meters.

Convert to and plug in values:

$\overrightarrow{E}=9.0*10^9\frac{-38*10^{-9}}{.015^2}$

$\overrightarrow{E}=-1.52*10^6\frac{N}{C}$

Magnitude is equivalent to absolute value:

$|\overrightarrow{E}|=1.52*10^6\frac{N}{C}$

Report an Error

Example Question #24 : Electrostatics

Charge A and B are $15\textup{ cm}$ apart. If charge A has a charge of $2\textup{ nC}$ and a mass of $2\textup{ mg}$ , charge B has a charge of $4\textup{ nC}$ and a mass of $10\textup{ mg}$ , determine the electric field at A due to B.

Possible Answers:

$537\ \frac{\textup{N}}{\textup{C}}$

$925\ \frac{\textup{N}}{\textup{C}}$

None of these

$1600\ \frac{\textup{N}}{\textup{C}}$

$611\ \frac{\textup{N}}{\textup{C}}$

Correct answer:

$1600\ \frac{\textup{N}}{\textup{C}}$

Explanation:

Using electric field formula:

$E=k\frac{q}{r^2}$

Converting to , to and plugging in values:

$E=9*10^9 \frac{4*10^{-9}}{.15^2}$

$E=1600 \frac{N}{C}$

Report an Error

Example Question #25 : Electrostatics

Possible Answers:

$667\ \frac{\textup{N}}{\textup{C}}$

None of these

$997\ \frac{\textup{N}}{\textup{C}}$

$118\ \frac{\textup{N}}{\textup{C}}$

$800\ \frac{\textup{N}}{\textup{C}}$

Correct answer:

$800\ \frac{\textup{N}}{\textup{C}}$

Explanation:

Using electric field formula:

$E=k\frac{q}{r^2}$

Converting to , to and plugging in values:

$E=9*10^9 \frac{2*10^{-9}}{.15^2}$

$E=800\frac{N}{C}$

Report an Error

Example Question #21 : Electrostatics

Two electrons are deep in space and apart. Determine the magnitude of the electric field at one electron due to the other.

Possible Answers:

$1.78*10^{-9}\frac{N}{C}$

$3.56*10^{-9}\frac{N}{C}$

None of these

$1.44*10^{-9}\frac{N}{C}$

$1.02*10^{-9}\frac{N}{C}$

Correct answer:

$1.44*10^{-9}\frac{N}{C}$

Explanation:

Using

$E=k\frac{q}{r^2}$

Plugging in values:

$E=9.0*10^9\frac{({-1.6*10^{-19}})}{1^2}$

$E=-1.44*10^{-9}\frac{N}{C}$

$|E|=1.44*10^{-9}\frac{N}{C}$

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All AP Physics 2 Resources

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AP Physics 2 : AP Physics 2

Study concepts, example questions & explanations for AP Physics 2

All AP Physics 2 Resources

Example Questions

Example Question #1 : Electric Fields

Example Question #18 : Electrostatics

Example Question #1 : Electric Fields

Example Question #2 : Electric Fields

Example Question #11 : Electric Fields

Example Question #11 : Electric Fields

Example Question #11 : Electric Fields

Example Question #24 : Electrostatics

Example Question #25 : Electrostatics

Example Question #21 : Electrostatics

All AP Physics 2 Resources

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