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AP Physics 2 : Electric Fields

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

Example Questions

Example Question #21 : Electrostatics

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:

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

None of these

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

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

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

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:

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

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Example Question #22 : Electrostatics

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:

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

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

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

None of these

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

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:

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

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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:

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

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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:

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

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

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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:

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

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

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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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Example Question #22 : Electrostatics

Knowing that the electric field $5\: m$ away from a point charge is $12\: \frac{N}{C}$ , what is the electric field at a distance of $15\:m$ away from this charge?

Possible Answers:

$2.25\: \frac{N}{C}$

$8.00\ \frac{N}{C}$

$5.00\: \frac{N}{C}$

$1.33\: \frac{N}{C}$

$3.50\: \frac{N}{C}$

Correct answer:

$1.33\: \frac{N}{C}$

Explanation:

In this question, we are given the strength of an electric field at a given distance away from a source charge. Then, we're asked to find what the strength of this electric field would be at a different distance away from the source charge.

First, we will need to use the equation for the electric field.

$E=\frac{kq}{r^{2}}$

Since we're given the strength of the electric field at a distance of $5\: m$ away, we can use this information to solve for the value of .

$E_{5}=\frac{kq}{(5\: m)^{2}}=12\: \frac{N}{C}$

$kq=(12\: \frac{N}{C})(5\: m)^{2} =300\: \frac{N\cdot m^{2}}{C}$

Now that we have the value for , we can use it to find the value of the electric field $15\: m$ from the source charge.

$E_{15}=\frac{kq}{(15\: m)^{2}}=\frac{300\: \frac{N\cdot m^{2}}{C}}{(15\: m)^{2}}=1.33\: \frac{N}{C}$

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Example Question #11 : Electric Fields

A charge is isolated deep in space. Determine the magnitude and direction of the electric field 10km away.

Possible Answers:

$3*10^{-9}\frac{N}{C}$ away from the charge

$9*10^{-8}\frac{N}{C}$ away from the charge

$9*\frac{N}{C}$ away from the charge

$9*10^{-8}\frac{N}{C}$ towards the charge

$9*\frac{N}{C}$ towards the charge

Correct answer:

$9*10^{-8}\frac{N}{C}$ away from the charge

Explanation:

Using Coulomb's law:

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

Where

$k=9.0*10^9\frac{N*m^2}{C^2}$

is the distance, in meters,

and is the charge, in Coulombs.

Plugging in values:

$E=(9*10^9)\frac{1*10^{-9}}{10,000^2}$

$E=9*10^{-8}\frac{N}{C}$

The positive sign indicates that the field is pointing away from the charge.

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All AP Physics 2 Resources

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AP Physics 2 : Electric Fields

Study concepts, example questions & explanations for AP Physics 2

All AP Physics 2 Resources

Example Questions

Example Question #21 : Electrostatics

Example Question #22 : Electrostatics

Example Question #11 : Electric Fields

Example Question #24 : Electrostatics

Example Question #25 : Electrostatics

Example Question #21 : Electrostatics

Example Question #22 : Electrostatics

Example Question #11 : Electric Fields

All AP Physics 2 Resources

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