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AP Physics 2 : Electricity and Magnetism

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Example Questions

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 #101 : Electricity And Magnetism

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

Possible Answers:

It will increase in magnitude and have constant direction

It will flip directions constantly

None of these

It will increase in magnitude and flip directions

It will decrease in magnitude and have constant direction

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 #751 : Ap Physics 2

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 #102 : Electricity And Magnetism

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

Impossible to determine

Positive work done

Infinite work done

Negative 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:

10N

80N

There is no way to determine the new force

40N

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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Example Question #3 : Electric Force Between Point Charges

You have two point charges, q_1 and q_2 , 3m away from each other. The value of q_1 is $3\mu C$ , and the force they both experience is 0.024 N . What is the value of q_2 ?

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

Possible Answers:

$5\mu C$

$16\mu C$

$6\mu C$

$12\mu C$

$8\mu C$

Correct answer:

$8\mu C$

Explanation:

The equation for the force between two point charges is as follows:

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

We have the values for , q_1 , , and , so we just need to rearrange the equation to solve for q_2 , then plug in the values we have.

$q_2=\frac{F\cdot r^2}{k\cdot q_1}$

$q_2= \frac{0.024 \cdot 3^2}{(9\cdot 10^9)\cdot (3\cdot 10^{-6})}$

$q_2= 8\cdot 10^{-6}$

Therefore, the value for the second charge is $8\mu C$ .

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Example Question #1 : Electric Force Between Point Charges

Two objects in space are placed 35m apart. One has a charge of 90e , the second has a charge of -95e . What is the electric force between them?

Possible Answers:

$1.6\cdot 10^{-27}N$ towards each other

$3.75\cdot 10^{-27}N$ towards each other

$1.6\cdot 10^{-27}N$ away from each other

$1.6\cdot 10^{27}N$ towards each other

$3.75\cdot 10^{-27}N$ away from each other

Correct answer:

$1.6\cdot 10^{-27}N$ towards each other

Explanation:

Use Coloumb's law:

$E_{force}=\frac{kq_1q_2}{r^2}$

First, convert the charges into coulombs.

$1 e = 1.6\cdot 10^{-19} C$

$q_1=1.44\cdot 10^{-17}C$

$q_2=-1.52\cdot 10^{-17}$

Plug in values into Coulomb's law.

$E_{force}=9\cdot10^{9}\frac{N\cdot m^{2}}{C^{2}}\cdot \frac{(1.44\cdot10^{-17}C)(-1.52\cdot10^{-17}C)}{(35m)^{2}}$

$E_{force}=-1.6\cdot 10^{-27}N$

Negative electric forces indicate attractive forces. Alternatively, we can reason that since one charge is positive, and the other is negative, the charges will attract.

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Example Question #756 : Ap Physics 2

Coloumbs law for varsity

If charge has a value of $-9*10^{-13}C$ , and charge has a value of $8*10^{-13}C$ , and the distance between their centers, , is 23 cm what will be the magnitude of the force on charge ?

Possible Answers:

$|F|=1.84*10^{-13}N$

None of these

$|F|=1.22*10^{-13}N$

$|F|=4.22*10^{-13}N$

$|F|=2.33*10^{-13}N$

Correct answer:

$|F|=1.22*10^{-13}N$

Explanation:

Using coulombs law to solve

$F=k\frac{q_1q_2}{r^2}$

Where:

q_1 it the first charge, in coulombs.

q_2 is the second charge, in coulombs.

is the distance between them, in meters

is the constant of $9*10^9\frac{N*m^2}{C^2}$

Converting into

$23cm*\frac{1m}{100cm}=.23m$

Plugging values into coulombs law

$F=9*10^{9}\frac{-9*10^{-13}8*10^{-13}}{.23^2}$

$F=-1.22*10^{-13}N$

The magnitude will be the absolute value.

$|F|=1.22*10^{-13}N$

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Example Question #1 : Electric Force Between Point Charges

Coloumbs law for varsity

If charge has a value of $-9*10^{-13}C$ , and charge has a value of $8*10^{-13}C$ , and the distance between their centers, , has a value of 66 cm , what will be the force on charge ?

Possible Answers:

None of these

$|F|=1.49*10^{-14}N$

$|F|=2.22*10^{-14}N$

$|F|=3.75*10^{-14}N$

|F|=1.49N

Correct answer:

$|F|=1.49*10^{-14}N$

Explanation:

Using coulombs law to solve

$F=k\frac{q_1q_2}{r^2}$

Where:

q_1 it the first charge, in coulombs.

q_2 is the second charge, in coulombs.

is the distance between them, in meters

is the constant of $9*10^9\frac{N*m^2}{C^2}$

Converting into

$66cm*\frac{1m}{100cm}=.66m$

Plugging values into coulombs law

$F=9*10^9\frac{(-9*10^{-13})(8*10^{-13})}{{.66}^2}$

$F=-1.49*10^{-14}N$

Magnitude will be the absolute value

$|F|=1.49*10^{-14}N$

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AP Physics 2 : Electricity and Magnetism

Study concepts, example questions & explanations for AP Physics 2

All AP Physics 2 Resources

Example Questions

Example Question #101 : Electricity And Magnetism

Example Question #101 : Electricity And Magnetism

Example Question #751 : Ap Physics 2

Example Question #102 : Electricity And Magnetism

Example Question #1 : Electric Force Between Point Charges

Example Question #1 : Electric Force Between Point Charges

Example Question #3 : Electric Force Between Point Charges

Example Question #1 : Electric Force Between Point Charges

Example Question #756 : Ap Physics 2

Example Question #1 : Electric Force Between Point Charges

All AP Physics 2 Resources

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