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AP Physics 2 : Electrostatics

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Example Questions

Example Question #71 : Electrostatics

A 30nC charge is located at the origin. A 50nC charge is located at (10,0)mm . Where could a proton be placed where it would experience no net force?

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

Possible Answers:

(8.74,0)mm

None of these

(2.19,0)mm

(4.37,0)mm

(0,4.37)mm

Correct answer:

(4.37,0)mm

Explanation:

Since both point charges are located on the x-axis, this is a one dimensional problem.

The proton, which has a positive charge, will be repelled by both charges.

Placing the proton somewhere "in the middle" will allow the forces to balance out.

$F_{net}=0=k\frac{q_1q_e}{r_1^2}-k\frac{q_2q_e}{r_2^2}$

Where

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

q_1 is the value of the first charge

q_2 is the value of the second charge

q_e is the charge of the proton

r_1 is the distance from the first charge to the proton

r_2 is the distance from the second charge to the proton

q_1 is 5mm away from q_2 , so:

r_1+r_2=10mm

Combining equations and plugging in values:

$F_{net}=0=9*10^9\frac{30*10^{-9}*1.6*10^{-19}}{r_1^2}-9*10^9\frac{50*10^{-9}*1.6*10^{-19}}{r_2^2}$

$\frac{4.32*10^{-17}}{r_1^2}=\frac{7.2*10^{-17}}{r_2^2}$

$\frac{r_1^2}{4.32*10^{-17}}=\frac{r_2^2}{7.2*10^{-17}}$

$\frac{r_1^2}{4.32}=\frac{r_2^2}{7.2}$

$r_1=\sqrt{\frac{4.32r_2^2}{7.2}}$

r_1=.775r_2

.775r_2+r_2=10mm

r_2=5.63mm

r_1=4.37mm

Thus, the location of balanced forces is at:

(4.37,0)mm

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Example Question #72 : Electrostatics

An electron is 500mm away from a point charge. It experiences a force of $2.6*10^{-16}N$ towards the point charge. Determine the value of the point charge.

Possible Answers:

$4.5*10^{-8}C$

$-1.6*10^{-19}C$

$9*10^{-8}C$

$4.5*10^{-1}C$

None of these

Correct answer:

$4.5*10^{-8}C$

Explanation:

Using

$F=k\frac{q_1q_e}{r^2}$

Solving for q_1

$F\frac{r^2}{k*q_e}=q_1$

Converting to and plugging in values

$-2.6*10^{-16}\frac{.500^2}{(9*10^{9})*-1.6*10^{-19}}=q_1$

*Note: a negative sign is used for the force because it is an attractive force, if it was a repulsive force, the opposite sign would be used.

$q=4.5*10^{-8}C$

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Example Question #21 : Electric Force Between Point Charges

Which of the following will cut the magnitude of the electric force between two charged particles in half?

Possible Answers:

Placing the two particles twice as far apart

Double the charge of only one of the particles

Reducing the charge of only one of the particles by half

Doubling the charge of each particle

Reducing the charge of each particle by half

Correct answer:

Reducing the charge of only one of the particles by half

Explanation:

To answer this question, it's necessary to understand the factors that affect the electric force. To show this, we can write the electrical force expression.

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

Using the above expression, we can look at how each answer choice would change the electric force.

If we place the two particles twice as far apart, the magnitude of the electric force will be reduced by a factor of .

Doubling the charge of one particle would double the electric force. Doubling the charge on both particles would cause the electric force to become times as great.

Halving the charge of both particles would cause the electric force to decrease by a factor of .

If the charge on only one of the particles is cut in half, then the electric force would be cut in half as well. Thus, this is the correct answer.

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Example Question #22 : Electric Force Between Point Charges

Two charges are a fixed distance apart. Both charges have charge . If another charge of charge and mass is placed a distance $\frac{d}{4}$ from one of the charges and $\frac{3d}{4}$ from the other, what will be the magnitude of its acceleration the moment it's released?

Possible Answers:

$\frac{16kqQ}{md^2}$

$\frac{8kqQ}{3md}$

$\frac{8kqQ}{9md}$

$\frac{128kqQ}{9md^2}$

$\frac{16kqQ}{9md^2}$

Correct answer:

$\frac{128kqQ}{9md^2}$

Explanation:

All we have to do is find the sum of the forces on the charge and divide by its mass. To find the force from each charge, we can use Coulomb's law:

$F=\frac{kq_{1}q_{2}}{d^2}$

Let's let the force from the charge a distance $\frac{d}{4}$ away be positive. That force is $\frac{kqQ}{(\frac{d}{4})^2}=\frac{16kqQ}{d^2}$ . The other force will be negative because it's acting in the opposite direction. This force is $\frac{-kqQ}{(\frac{3d}{4})^2}=\frac{-16kqQ}{9d^2}$ . Adding these two together we get

$\frac{128kqQ}{9d^2}$

This is the magnitude of the net force. To find acceleration, we divide by the mass to get

$\frac{128kqQ}{9md^2}$

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Example Question #1 : Electric Potential Energy

$6.1 \cdot 10^{20}$ electrons pass through a $20\Omega$ resistor in 11 minutes. What is the potential difference across that resistor?

$e=1.602\cdot 10^{-19}$

Possible Answers:

8.11V

4.45V

1.34V

6.12V

2.96V

Correct answer:

2.96V

Explanation:

We need to find the voltage, and we have the resistance, so if we can find the current, then we can use Ohm's Law to find the voltage.

The definition of current is amount of charge that flows through a point in time, so current can be calculated using this equation:

$I=\frac{\Delta Q}{\Delta t}$

We're told how many electrons have passed through a resistor, and we know how much charge a single electron has. If we convert the number of electrons into total amount of charge, we can divide that number by the amount of time in seconds to find the current.

$6.1\cdot 10^{20} e^-\cdot 1.602\cdot 10^{-19}\frac{C}{electron}=97.722 C$

$11min\cdot60\frac{s}{min}=660s$

Now we have the amount of charge in and the amount of time in seconds, so we divide the two.

$I=\frac{\Delta Q}{\Delta t}$

$I=\frac{97.722}{660}$

I= 0.1481

Now that we have the current, we can use Ohm's Law to find the voltage.

V=IR

V=(0.1481)(20)

V= 2.961

Therefore, the potential difference is 2.96V.

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Example Question #1 : Electric Potential Energy

What is the electric potential between two terminals of a cell if it requires $\textup {7J}$ of work to transfer $\textup {3C}$ between the terminals?

Possible Answers:

$\frac{3}{7}\textup{eV}$

$\frac{3}{7}\textup{V}$

$\frac{7}{3}\textup{V}$

$\frac{7}{3}\textup{A}$

$\frac{3}{49}\textup{A}$

Correct answer:

$\frac{7}{3}\textup{V}$

Explanation:

Write the formula for the potential difference.

$V=\frac{W}{Q}$

Substitute the work and charge. The unit is in volts.

$V=\frac{W}{Q}= \frac{7\textup{J}}{3\textup{C}}=\frac{7}{3}\textup{V}$

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Example Question #1 : Electric Potential Energy

There is an electric field of between two parallel plates of $3.5\cdot 10^{-5}\frac{N}{C}$ . The two plates are 3.5mm apart.

Find the electric potential energy of a particle of charge 2.75nC placed right at the surface of the positive plate.

Possible Answers:

$7\cdot10^{-16} J$

$3.37\cdot 10^{-16}J$

None of these

$9.65\cdot 10^{-16} J$

$2.33\cdot10^{-16} J$

Correct answer:

$3.37\cdot 10^{-16}J$

Explanation:

Use the equations for potential electric energy and electric potential:

$PE_{electric} = Vq$

V=Ed

Substitute.

$PE_{electric} = Edq$

Plug in known values and solve.

$PE_{electric}= (2.75\cdot 10^{-9}C)(35\cdot 10^{-6}\frac{N}{C})(3.5\cdot 10^{-3}m)$

$PE_{electric} = 337\cdot 10^{-18} Nm = 337\cdot10^{-18} J=3.37\cdot 10^{-16}J$

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Example Question #2 : Electric Potential Energy

A student is working on a laboratory exercise in which she measures the electric potential (voltage) at several points in an electric field. Before graphing the potential isolines, she measures the electric potential to be 3V at one point in the field and 7V at a point 3cm from the first point. What is the average electric field strength between the two points the student measured?

Possible Answers:

$233\frac{N}{C}$

$133\frac{N}{C}$

$100\frac{N}{C}$

$2.33\frac{N}{C}$

$1.33\frac{N}{C}$

Correct answer:

$133\frac{N}{C}$

Explanation:

Electric field strength is the slope of the electric potential:

$E=\frac{\Delta V}{\Delta x}$

Remember to convert into SI units (meters):

$E=\frac {V_2-V_1}{\Delta x}$

$E=\frac{7V-3V}{0.03m}$

$E=133 \frac{V}{m} = 133 \frac{N}{C}$

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Example Question #2 : Electric Potential Energy

In a region of space, there is an electric field whose magnitude is $E=20\frac{N}{C}$ pointing due North. An electron enters this field traveling due North with an initial velocity $v=2.4\cdot10^{4}\frac{m}{s}$ . It enters the field at point A, where the electric potential is 1.5V. As it travels 2cm in the field to point B, the potential changes to 0V. What will the electron's velocity be when it arrives at point B?

$m=9.11\cdot10^{-31}kg$

$q=-1.6\cdot10^{-19}C$

Possible Answers:

$3.6\cdot 10^{4}\frac{m}{s}$

$4.8\cdot 10^{5}\frac{m}{s}$

$3.63\cdot 10^{5}\frac{m}{s}$

$1.6\cdot 10^{5}\frac{m}{s}$

Cannot be determined

Correct answer:

$3.63\cdot 10^{5}\frac{m}{s}$

Explanation:

This problem is solved using the work-kinetic energy theory: $\Sigma W=\Delta K$ . In an electric field, work is equal to charge times change in potential: $W=q\Delta V$

Since an electron has a negative charge, decreasing potential increases its kinetic energy, the opposite of what would happen to a proton with its positive charge. Combine these equations:

$q\Delta V=K_f-K_i$

$q\Delta V=\frac{1}{2}mv_f^{2}-\frac{1}{2}mv_i^{2}$

Plug in known values and solve

$1.6\times10^{-19}\cdot 1.5=\frac{1}{2}\cdot 9.11\cdot 10^{-31}v_f^{2}-\frac{1}{2}\cdot9.11\cdot10^{-31}\cdot(2.4\cdot10^{4})^{2}$

$v_f=3.63\cdot 10^{5}\frac{m}{s}$

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Example Question #6 : Electric Potential Energy

Suppose that a point charge of 1 Coulomb undergoes a change in which it is moved from point A to point B while in the presence of an external electric field. During this transposition, it undergoes a voltage change of $+6\: Volts$ . What change in electrical potential energy occurs in this scenario?

Possible Answers:

0.083 J

0.167 J

12 J

Correct answer:

Explanation:

In order to solve for electrical potential energy, we'll need to remember the equation for it.

$U=k\frac{q_{s}q_{t}}{r}$

In the above expression, represents electrical potential energy, $q_{s}$ and $q_{t}$ represent different point charges, and represents the distance between their centers. In this example, one of these charges will be the source of the external electric field, while the other charge will be the one that is undergoing a transposition from point A to point B.

Furthermore, we can remember the equation for voltage:

$V=k\frac{q_{s}}{r}$

With both these equations in mind, we can combine the two:

$U=Vq_{t}$

This above expression tells us that the electrical potential energy of a system is directly proportional to the voltage change and to the charge of the test charge that is undergoing the voltage change.

Plug in the values and solve for electrical potential energy:

U=(6V)(1 C)=6J

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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 #71 : Electrostatics

Example Question #72 : Electrostatics

Example Question #21 : Electric Force Between Point Charges

Example Question #22 : Electric Force Between Point Charges

Example Question #1 : Electric Potential Energy

Example Question #1 : Electric Potential Energy

Example Question #1 : Electric Potential Energy

Example Question #2 : Electric Potential Energy

Example Question #2 : Electric Potential Energy

Example Question #6 : Electric Potential Energy

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