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AP Physics 1 : Electrostatics

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Example Questions

Example Question #2 : 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:

0.89876N

26.9628N

13.4814N

8.9876N

2.69628N

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 #91 : Electricity And Waves

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

Possible Answers:

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

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

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

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

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 #91 : Electricity And Waves

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:

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

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

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

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

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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Example Question #1 : Fundamentals Of Electric Charge

A point charge of charge $q = 45\mu C$ is placed in a uniform electric field of strength $78 \frac{N}{C}$ going to the right. The charge is then moved at an angle of $30^{\text o}$ below the horizontal to the right. What is the net work done on this charge?

Possible Answers:

$-1.56\cdot10^{-3}J$

$3.04\cdot10^{-3}J$

$2.36\cdot10^{-3}J$

$-6.04\cdot10^{-3}J$

$1.76\cdot10^{-3}J$

Correct answer:

$3.04\cdot10^{-3}J$

Explanation:

The work done by an electric field on a charged particle is:

$W=qEcos(\theta)$

Where $\theta$ is the angle between the electric field vector and the displacement vector. Plug in the given values.

$3.04\cdot10^{-3}J$

This answer makes sense since an electric field does work on a positive charge when going in the direction of the field lines.

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Example Question #2 : Fundamentals Of Electric Charge

Two charges of +2q and +5q are placed 0.5m away from each other. Another charge is placed in between these two charges. Where should this third charge be placed in order for it to feel no net electrostatic force?

Possible Answers:

0.17 m from the +2q charge

There is not enough information given in this problem to solve the question

0.17 m from the +4q charge

0.21 m from the +4q charge

0.21 m from the +2q charge

Correct answer:

0.21 m from the +2q charge

Explanation:

The electrostatic force between two charges is:

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

The force that the 3rd charge feels from the +2q charge is:

$F = k\frac{2qq_{3}}{r^{2}_{13}}$

The force that the 3rd charge feels from the +4q charge is:

$F = k\frac{4qq_{3}}{r^{2}_{23}}$

We can also rewrite the distance between the charges as and 0.5-x . In this explanation, we will use as the distance between the +2q and the third unknown charge making the distance between the +4q and the 3rd charge 0.5-x . Finally, we solve for a net force of 0 giving us this relationship between the two opposing forces.

$k\frac{2qq_{3}}{x^{2}} = k\frac{4qq_{3}}{(.5 - x)^{2}}$

Solve for .

$\frac{2}{x^{2}} = \frac{4}{(.5 - x)^{2}}$

x = 0.21m

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Example Question #1 : Fundamentals Of Electric Charge

A student has a neutrally charged glass rod and a neutrally charged silk cloth. When the student rubs the silk cloth on the glass rod, the rod acquires a net positive charge. What is the charge on the silk cloth after the student performs this experiment?

Possible Answers:

The silk cloth has no charge

The silk cloth has the same charge as the glass rod

The silk cloth is negatively charged and has a smaller magnitude of charge than the glass rod

The silk cloth has a charge that is equal in magnitude to the glass rod's charge, but is negative

The silk cloth is positively charged and has a greater magnitude of charge than the glass rod.

Correct answer:

The silk cloth has a charge that is equal in magnitude to the glass rod's charge, but is negative

Explanation:

Charge is fundamentally conserved. In order to give the glass rod a positive charge, the silk cloth removes electrons from the glass rod. Every electron that leaves the glass rod leaves behind the proton (positively charged particle) that used to balance the negative charge of the electron (negatively charged particle). Every electron that leaves the glass rod goes onto the silk cloth, giving it the same amount of charge as the glass rod, but with a negative sign (since electrons are negatively charged). Only a tiny fraction of the total electrons in the glass rod are removed, even with vigorous charging.

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Example Question #1251 : Ap Physics 1

If the test charge is increased by a factor of 4, what happens to the electric potential of that charge?

Possible Answers:

Decreases by a factor of

Remains unchanged

Increases by a factor of

Correct answer:

Remains unchanged

Explanation:

Remember that the formula for the electric potential is given by:

$V=\frac{kQ}{r}$

Where is Coulomb's constant, is the source charge, and is the distance. This formula indicates that electric potential is unaffected by changing the test charge.

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AP Physics 1 : Electrostatics

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #2 : Coulomb's Law

Example Question #3 : Coulomb's Law

Example Question #21 : Electrostatics

Example Question #91 : Electricity And Waves

Example Question #91 : Electricity And Waves

Example Question #13 : Coulomb's Law

Example Question #1 : Fundamentals Of Electric Charge

Example Question #2 : Fundamentals Of Electric Charge

Example Question #1 : Fundamentals Of Electric Charge

Example Question #1251 : Ap Physics 1

All AP Physics 1 Resources

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