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AP Physics 2 : Electric Force in an Electric Field

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Example Question #11 : Electric Force In An Electric Field

Calculate the force experienced by a particle with charge of q=3e^+ in the presence of an electric field $\vec{E}=500\hat{y}\left( \frac{N}{C} \right)$

$e=1.602*10^{-19}C$

Possible Answers:

$2.40\times 10^{-16}(N)$

$2.40\times 10^{-16}\hat{y}(N)$

$-2.40\times 10^{-16}\hat{z}(N)$

$2.40\times 10^{-16}\hat{x}(N)$

$-2.40\times 10^{-16}\hat{y}(N)$

Correct answer:

$2.40\times 10^{-16}\hat{y}(N)$

Explanation:

The force $\vec{F}$ on a given charge in the presence of an electric field $\vec{E}$ is given below as:

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

It is important to note that the force will be directed along the same direction as the electric field. Therefore, for our problem, we are able to write

$\vec{F}=\left(3e^+ \right )\left(500\hat{y} \right )\left(\frac{N}{C} \right )$

$\vec F=3\left(1.602 * 10^{-19}C \right )500 \hat{y} \frac{N}{C}=2.40 \times 10^{-16}\hat{y}(N)$

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Example Question #12 : Electric Force In An Electric Field

A test charge of $6\mu C$ is placed in an electric field of $<.005,.005,.003>\frac{N}{C}$ . Determine the force on the charge.

Possible Answers:

None of these

$\overrightarrow{F}=<6*10^{-8},4*10^{-8}, 7.8*10^{-8}>N$

$\overrightarrow{F}=<9*10^{-8},9*10^{-8}, 1.8*10^{-8}>N$

$\overrightarrow{F}=<3*10^{-8},3*10^{-8}, 1.8*10^{-8}>N$

$\overrightarrow{F}=<3*10^{-8},3*10^{-8}, 4*10^{-8}>N$

Correct answer:

$\overrightarrow{F}=<3*10^{-8},3*10^{-8}, 1.8*10^{-8}>N$

Explanation:

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

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

$\overrightarrow{F}=<.005,.005,.003>\frac{N}{C}*(6*10^{-6}C)$

$\overrightarrow{F}=<3*10^{-8},3*10^{-8}, 1.8*10^{-8}>N$

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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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AP Physics 2 : Electric Force in an Electric Field

Study concepts, example questions & explanations for AP Physics 2

All AP Physics 2 Resources

Example Questions

Example Question #11 : Electric Force In An Electric Field

Example Question #12 : Electric Force In An Electric Field

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

All AP Physics 2 Resources

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