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AP Physics 1 : Coulomb's Law

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Example Question #11 : Coulomb's Law

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:

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

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

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

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

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 #12 : Coulomb's Law

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

Possible Answers:

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

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

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

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

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 #13 : Coulomb's Law

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:

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

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

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

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

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 #11 : 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:

$230*10^{15}N$

$230*10^{-15}N$

230N

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 : Coulomb's Law

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

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Example Question #11 : Coulomb's Law

Example Question #12 : Coulomb's Law

Example Question #13 : Coulomb's Law

Example Question #11 : Coulomb's Law

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