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AP Physics 2 : Electrostatics

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Example Questions

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Example Question #13 : Electric Force In An Electric Field

A test charge of -9nC is placed in an electric field of $<-.008,-.008,.003>\frac{N}{C}$ . Determine the force on the charge.

Possible Answers:

None of these

$\overrightarrow{F}=<7.2*10^{-11},7.2*10^{-11}, -2.7*10^{-11}>N$

$\overrightarrow{F}=<6.9*10^{-11},7.2*10^{-11}, 3.8*10^{-11}>N$

$\overrightarrow{F}=<7.2*10^{-11},4.5*10^{-11}, 4.5*10^{-11}>N$

$\overrightarrow{F}=<-7.2*10^{-11},-7.2*10^{-11}, 2.7*10^{-11}>N$

Correct answer:

$\overrightarrow{F}=<-7.2*10^{-11},-7.2*10^{-11}, 2.7*10^{-11}>N$

Explanation:

The definition of electric force on a charge in an electric field is:

$\overrightarrow{F}=\overrightarrow{E}q$

$\overrightarrow{F}=<-.008,-.008,.003>\frac{N}{C}*(9*10^{-9}C)$

$\overrightarrow{F}=<-7.2*10^{-11},-7.2*10^{-11}, 2.7*10^{-11}>N$

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Example Question #14 : Electric Force In An Electric Field

A test charge of 35nC is placed in an electric field of $<.007,.006,.019>\frac{N}{C}$ . Determine the force on the charge.

Possible Answers:

$\overrightarrow{F}=<2.45*10^{-10},2.1*10^{-10}, 6.65*10^{-10}>N$

None of these

$\overrightarrow{F}=<6.77*10^{-10},2.3*10^{-10}, 8.7*10^{-10}>N$

$\overrightarrow{F}=<3.11*10^{-10},5.25*10^{-10}, 3.11*10^{-10}>N$

$\overrightarrow{F}=<2.10*10^{-10},1.38*10^{-10}, 9.87*10^{-10}>N$

Correct answer:

$\overrightarrow{F}=<2.45*10^{-10},2.1*10^{-10}, 6.65*10^{-10}>N$

Explanation:

The definition of electric force on a charge in an electric field is:

$\overrightarrow{F}=\overrightarrow{E}q$

$\overrightarrow{F}=<.007,.006,.019>\frac{N}{C}*(35*10^{-9}C)$

$\overrightarrow{F}=<2.45*10^{-10},2.1*10^{-10}, 6.65*10^{-10}>N$

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Example Question #15 : Electric Force In An Electric Field

Determine the force on a point charge of q=-33nC in an electric field of $<2.5,3.1>\frac{N}{C}$

Possible Answers:

None of these

$<7.77*10^{-9},1.02*10^{-7}>N$

$<-1.25*10^{-9},-8.02*10^{-7}>N$

$<8.25*10^{-9},1.02*10^{-7}>N$

$<-8.25*10^{-9},-1.02*10^{-7}>N$

Correct answer:

$<-8.25*10^{-9},-1.02*10^{-7}>N$

Explanation:

Use the following form of the electric force equation:

F_E=Eq

Plug in values:

$F_E=-33*10^{-9}*<2.5,3.1>\frac{N}{C}$

$F_E=<-8.25*10^{-9},-1.02*10^{-7}>N$

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Example Question #16 : Electric Force In An Electric Field

An object deep in space of mass 50mg and charge 60nC accelerates at $1.1\frac{m}{s^2}$ . Determine the electric field at this location.

Possible Answers:

$8.44*10^3\frac{N}{C}$

$4.49*10^3\frac{N}{C}$

$1.11*10^3\frac{N}{C}$

$9.17*10^2\frac{N}{C}$

$3.86*10^2\frac{N}{C}$

Correct answer:

$9.17*10^2\frac{N}{C}$

Explanation:

Use the following equation for electric force:

F_E=Eq

Plug in values:

$50*10^{-6}*1.1=60*10^{-9}*E$

$E=9.16*10^3\frac{N}{C}$

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Example Question #101 : Electricity And Magnetism

A $95\textup{ nC}$ charge of mass $55\ \mu\textup{g}$ accelerates with $<50,0>\ \frac{\textup{m}}{\textup{s}^2}$ . Determine the electric field.

Possible Answers:

$E=<48,0>\ \frac{\textup{N}}{\textup{C}}$

$E=<12,9>\ \frac{\textup{N}}{\textup{C}}$

$E=<15,15>\ \frac{\textup{N}}{\textup{C}}$

$E=<0,0>\ \frac{\textup{N}}{\textup{C}}$

$E=<28.9,0>\ \frac{\textup{N}}{\textup{C}}$

Correct answer:

$E=<28.9,0>\ \frac{\textup{N}}{\textup{C}}$

Explanation:

Using F_E=Eq and F=ma

Combining equations:

Eq=ma

Solving for

$\frac{m*a}{q}=E$

Converting $\mu g$ to , to and plugging in values

$\frac{55*10^{-9}*<50,0>}{95*10^{-9}}=E$

$E=<28.9,0>\frac{N}{C}$

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Example Question #11 : Electric Force In An Electric Field

An observer is away from a negatively charged sphere. How will the electric field change as they move closer to the sphere?

Possible Answers:

None of these

It will increase in magnitude and have constant direction

It will increase in magnitude and flip directions

It will decrease in magnitude and have constant direction

It will flip directions constantly

Correct answer:

It will increase in magnitude and have constant direction

Explanation:

E field negative

As one moves closer to a negative charge, the electric field increases in the direction of the negative charge.

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Example Question #103 : Electricity And Magnetism

Suppose I have a uniform electric field within a parallel plate capacitor with field strength of $E=41\frac{N}{C}$ .

Suppose the capacitor's plates are .5m in length and the space between the plates is 1.5cm .

Determine the magnitude of force experienced by an individual electron placed in this field and the direction. Assume the charge of an electron is $-1.602*10^{-19}C$

Possible Answers:

$6.6*10^{-18}N$ towards the negative plate

$6.6*10^{-18}N$ towards the positive plate

41N towards the positive plate

$6.6*10^{-19}N$ towards the positive plate

Correct answer:

$6.6*10^{-18}N$ towards the positive plate

Explanation:

Recall that the formula for the magnitude of force is given by:

F=|E*q|

Where is the electric field strength and is the charge.

Since the electric field strength is constant,

$F=(41 \frac{N}{C})*(1.602*10^{-19}C)=6.6*10^{-18}N$

To determine direction, remember that electrons will move towards the positively charged plates since electrons are negatively charged.

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Example Question #751 : Ap Physics 2

An electron is moved in a perfect circle, with a negative point charge in the center. Determine the work done by the negative point charge.

Possible Answers:

No work done

Infinite work done

Impossible to determine

Negative work done

Positive work done

Correct answer:

No work done

Explanation:

There will be no work done. The point charge at the center is emitting an electric field towards itself. If the electron is moved in a circle around it, it will have moved perpendicular to the force the entire time, and thus no work has been done.

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Example Question #1 : Electric Force Between Point Charges

Two charges are placed a certain distance apart such that the force that each charge experiences is 20 N. If the distance between the charges is doubled, what is the new force that each charge experiences?

Possible Answers:

80N

40N

10N

There is no way to determine the new force

Correct answer:

Explanation:

To solve this problem, we'll need to utilize the equation for the electric force:

$F_{e}=k\frac{Q_{1}Q_{2}}{r^{2}}$

We're told that the force each charge experiences is 20 N at a certain distance, but then that distance is doubled. Thus, the new electric force will be:

$F_{e}^{'}=k\frac{Q_{1}Q_{2}}{(2r)^{2}}=k\frac{Q_{1}Q_{2}}{4r^{2}}=\frac{F_{e}}{4}=\frac{20N}{4}=5N$

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Example Question #1 : Electric Force Between Point Charges

What is the force experienced by a $20\mu C$ point charge away from a $30\mu C$ point charge?

Possible Answers:

The point charge experiences no force

0.216N

0.108N

1.08N

5.4N

Correct answer:

0.216N

Explanation:

The equation to find the force from two point charges is called Coulomb's Law.

$F=\frac{kq_1q_2}{r^2}$

In this equation, is force in Newtons, is the respective charge value in $\mu C$ , is radius in meters, and is the Coulomb constant, which has a value of $9 \cdot 10^9 \frac{N*m^2}{C^2}$ .

Now, we just plug in the numbers. Note: the charge values are in microcoulombs (the Greek letter $\mu$ , called "mu," stands for micro), which is equal to $10^{-6}$ .

$F&=\frac{kq_1q_2}{r^2}$

$F=\frac{(9\cdot 10^9\frac{N*m^2}{C^2}) \cdot (20\cdot 10^{-6}\mu C) \cdot (30 \cdot 10^{-6}\mu C)}{(5m)^2}$

F= 0.216

Therefore, the force experienced on either charge is 0.216N of force.

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AP Physics 2 : Electrostatics

Study concepts, example questions & explanations for AP Physics 2

All AP Physics 2 Resources

Example Questions

Example Question #13 : Electric Force In An Electric Field

Example Question #14 : Electric Force In An Electric Field

Example Question #15 : Electric Force In An Electric Field

Example Question #16 : Electric Force In An Electric Field

Example Question #101 : Electricity And Magnetism

Example Question #11 : Electric Force In An Electric Field

Example Question #103 : Electricity And Magnetism

Example Question #751 : Ap Physics 2

Example Question #1 : Electric Force Between Point Charges

Example Question #1 : Electric Force Between Point Charges

All AP Physics 2 Resources

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