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AP Physics 1 : AP Physics 1

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Example Questions

Example Question #3 : Coulomb's Law

If we have 2 charges, q_1 and q_2 , that are 50cm apart, what is the magnitude of the force exerted on q_1 by q_2 if we know that q_1 has a charge of $10\mu C$ and q_2 has a charge of $-20\mu C$ ?

Possible Answers:

7.2N

2.7N

6.3N

3.6N

Correct answer:

7.2N

Explanation:

Use Coulomb's Law

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

Plug in known values and solve.

$\frac{(9\cdot10^{9})(10\cdot10^{-6})(-20\cdot10^{-6})}{0.5^{2}}$

F=-7.2N

A negative value for electric force indicates an attractive force. This makes sense since our two charges have opposite signs. Since we're asked for magnitude, all answer choices are positive.

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Example Question #4 : Coulomb's Law

If we have 2 charges, q_1 and q_2 , that are 14cm apart, what is the force exerted on q_1 by q_2 if we know that q_1 has a charge of $-13\mu C$ and q_2 has a charge of $-12\mu C$ ?

Possible Answers:

71.63N

52.86N

67.31N

143.16N

Correct answer:

71.63N

Explanation:

Use Coulomb's law.

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

Plug in known values and solve.

$F=\frac{(9\cdot10^{9})(-12\cdot10^{-6})(-13\cdot10^{-6})}{(0.14^{2})}$

F=71.63N

Note that this force is positive, which means it's repulsive.

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Example Question #5 : Coulomb's Law

If we have 2 charges, q_1 and q_2 , that are 10m apart, what is the force exerted on q_1 by q_2 if we know that q_1 has a charge of $15\mu C$ and q_2 has a charge of $20\mu C$ ?

Possible Answers:

270N

0.027N

27N

2.7N

Correct answer:

0.027N

Explanation:

Use Coulomb's law.

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

Plug in known values and solve.

$F=\frac{(9\cdot10^{9})(15\cdot10^{-6})(20\cdot10^{-6})}{(10^{2})}$

F=0.027N

Note that this force is positive, which means that it's repulsive.

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Example Question #5 : Coulomb's Law

If we have 2 charges, q_1 and q_2 , that are 15cm apart, what is the force exerted on q_1 by q_2 if we know that q_1 has a charge of $15\mu C$ and q_2 has a charge of $15\mu C$ ?

Possible Answers:

15N

90N

180N

45N

Correct answer:

90N

Explanation:

Use Coulomb's law.

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

Plug in known values and solve.

$F=\frac{(9\cdot10^{9})(15\cdot10^{-6})(15\cdot10^{-6})}{(0.15^{2})}$

F=90N

Note that the force between two charges of the same sign (both positive or both negative) is positive. This indicates the force is repulsive, which makes sense since both charges are positive.

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Example Question #1 : Coulomb's Law

Two point charges, $q_1=60\mu C$ and $q_2=150\mu C$ are separated by a distance of . What is the force of repulsion between them?

$k=8.9876\cdot10^9\frac{Nm^2}{C^2}$

Possible Answers:

8.9876N

26.9628N

0.89876N

2.69628N

13.4814N

Correct answer:

8.9876N

Explanation:

The force of attraction/repulsion between two point charges is given by Coulomb's Law:

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

If the charges are of like sign, then there well be a repulsive force between the two. Alternatively, if the net force is positive, it is repulsive; if it is negative, it is attractive.

Therefore, the force of repulsion between the two charges is:

$F=8.9876\cdot10^9\frac{Nm^2}{C^2}\cdot\frac{60\mu C(150\mu C)}{(3m)^2}$

$F=8.9876\cdot10^9\frac{Nm^2}{C^2}\cdot\frac{60\cdot10^{-6} C(150\cdot10^{-6} C)}{(3m)^2}$

$F=8.9876\cdot10^9\cdot\frac{Nm^2}{C^2}\cdot\frac{9000\cdot10^{-12} C^2)}{9m^2}$

F=8.9876N

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Example Question #3 : Coulomb's Law

Two point charges, $q_1=60\mu C$ and $q_2=150\mu C$ are separated by a distance of . What is the work required to move them closer together to a distance of ?

$k=8.9876\cdot10^9\frac{Nm^2}{C^2}$

Possible Answers:

13.48J

20.22J

40.44J

11.23J

8.99J

Correct answer:

13.48J

Explanation:

The force of attraction/repulsion between two point charges is given by Coulomb's Law:

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

If the charges are of like sign, then there well be a repulsive force between the two.

Work is given as the dot product of force and distance. However, in this case, force is also dependent on distance.

The amount of work required to move a charge an incremental distance, , is given as:

$dW=-k\frac{q_1q_2}{r^2}dr$

The negative sign in this case is to account for repulsion.

The total work to change distances between charges can then be found by taking the integral with respect to distance:

$W=\int_{d_1}^{d^2} -k\frac{q_1q_2}{r^2}dr$

Since k,q1,q2 are constants, they can be factored out of the integral:

$W=-kq_1q_2\int_{d_1}^{d_2} \frac{1}{r^2}dr$

$W=kq_1q_2\frac{1}{r}|_{d_1}^{d_2}$

$W=(8.9876\cdot10^9\frac{Nm^2}{C^2})(60\cdot10^{-6}C)(150\cdot10^{-6}C) \left(\frac{1}{2m}-\frac{1}{3m}\right)$

W=13.48J

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Example Question #21 : Electrostatics

Write, in vector notation, the force exerted on a positive charge of by a negative charge of -2C , if the two charges sitting on the -axis, with the positive charge sitting 100m to the right of the negative charge?

Possible Answers:

$\vec{F_e}=(0 N)\hat{i}+(-1.80\times10^6 N)\hat{j}$

$\vec{F_e}=(-3.6\times10^6 N)\hat{i}+(0N)\hat{j}$

$\vec{F_e}=(-1.80\times10^6 N)\hat{i}+(0N)\hat{j}$

$\vec{F_e}=(1.80\times10^6 N)\hat{i}+(0N)\hat{j}$

Correct answer:

$\vec{F_e}=(-1.80\times10^6 N)\hat{i}+(0N)\hat{j}$

Explanation:

Coulomb's law in vector notation is given as:

$F_{e}=\frac{kq_1q_2}{r^2}\hat{r}$ , where is Coulomb's constant, q_1 and q_2 are the two charges, r^2 is the distance between the charges squared, and $\hat{r}$ is the unit vector going from one charge to another.

To write this in vector notation, we have to know the unit vector going from the negative to the positive charge, since we're trying to determine the force on the positive charge. Since they are both sitting on the -axis, with the negative charge to the left of the positive, the unit vector will be going in the direction of positive :

$\hat{r}=<1,0>$

We know that $k=8.99\times10^9 Nm^2C^{-2}$

We know that r=100m , q_1=-2C , and q_2=1C . Putting this together:

$\vec{F_e}=\frac{8.99\times10^9Nm^2C^{-2} \cdot1C\cdot-2C}{(100\:m)^2}<1,0>$

$\vec{F_e}=-1.80*10^6N*<1,0>= <-1.80*10^6N,0N>$

We can rewrite this as:

$\vec{F_e}=(-1.80\times10^6N)\hat{i}+(0N)\hat{j}$

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Example Question #21 : Electrostatics

What are the unit(s) of Coulomb's constant ?

Possible Answers:

$\frac{N}{C^2m^2}$

$\frac{N\:m}{C^2}$

$\frac{N\:m^2}{C^2}$

$\frac{C^2\:m^2}{N}$

Correct answer:

$\frac{N\:m^2}{C^2}$

Explanation:

To determine this, we have to solve for in Coulomb's law and then determine its constants.

Recall that the magnitude of the electrostatic force between point charges is given as:

$F_e=\frac{kq_1q_2}{r^2}$ , F_e is the force given in , q_1 and q_2 are the charges given in and is distance given in

Solving for ,

$\frac{F_e\:r^2}{q_1q_2}=k$

Writing out the terms on the left in their units:

$\frac{N\:m^2}{C^2}=k$

Therefore, is given in $\frac{N\:m^2}{C^2}$

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Example Question #21 : Electrostatics

Two protons are at a distance $r_{1}$ away from each other. There is a force $F_{1}$ acting on each proton due to the other. If the protons are moved so that they are now at a distance

$r_{2}=\frac{1}{2} r_{1}$ apart, what is the new force acting on each proton due to the other $\left (F_{2} \right )$ ?

Possible Answers:

$F_{2}=\frac{1}{2} F_{1}$

$F_{2}=4 F_{1}$

$F_{2}=\frac{1}{4} F_{1}$

$F_{2}=2 F_{1}$

Correct answer:

$F_{2}=4 F_{1}$

Explanation:

Coulomb's law shows that the force between two charged particles is inversely proportional to the square of the distance between the particles.

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

If the distance between the charges is reduced by $\frac{1}{2}$ , that means the $\frac{1}{2}$ is squared in the denominator and the will flip up to the top to give time the original force. More explicitly, if we plug in the given information the initial force will be:

$F_{1}=k \frac{q_{1}q_{2}}\left ( {r_{1}\right )^{2}}$

$F_{2}=k \frac{q_{1}q_{2}}\left ( {r_{2}\right )^{2}}=k \frac{q_{1}q_{2}}\left ( \frac{1}{2}{r_{1}\right )^{2}}$

$F_{2}=k \frac{q_{1}q_{2}}{ \frac{1}{4}\left (r_{1}\right )^{2}}=4 k \frac{q_{1}q_{2}}{\left (r_{1}\right )^{2}}$

$F_{2}=4 F_{1}$

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Example Question #13 : Coulomb's Law

Determine the strength of a force of proton on another proton in the nucleus if they are $10^{-15}m$ apart.

$k=8.99*10^9\frac{N}{m^2C^2}$

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

Possible Answers:

230N

$230*10^{-15}N$

$230*10^{15}N$

57.5N

Correct answer:

230N

Explanation:

Use Coulomb's law:

$F_c=\frac{k*Q*q}{r^2}$ , where is Coulomb's constant, Q,q are charges of the two points and is the distance between the charges.

$F_c=\frac{8.99*10^9*(1.602*10^{-19})^2}{(10^{-15})^2}=230N$

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AP Physics 1 : AP Physics 1

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #3 : Coulomb's Law

Example Question #4 : Coulomb's Law

Example Question #5 : Coulomb's Law

Example Question #5 : Coulomb's Law

Example Question #1 : Coulomb's Law

Example Question #3 : Coulomb's Law

Example Question #21 : Electrostatics

Example Question #21 : Electrostatics

Example Question #21 : Electrostatics

Example Question #13 : Coulomb's Law

All AP Physics 1 Resources

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