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AP Physics C Electricity : Using Coulomb's Law

Study concepts, example questions & explanations for AP Physics C Electricity

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Example Questions

Example Question #1 : Electricity And Magnetism Exam

We have a point charge of 100nC . Determine the electric field at a distance of $500\mu m$ away from that charge.

$k=8.99* 10^9\ \frac{\text{Nm}^2}{\text{C}^2}$

Possible Answers:

$3.6*10^9\frac{N}{C}$

$2.7*10^8\frac{N}{C}$

$8.5*10^2\frac{N}{C}$

$3.6*10^6\frac{N}{C}$

$6.3*10^5\frac{N}{C}$

Correct answer:

$3.6*10^9\frac{N}{C}$

Explanation:

Coulomb's law for the electric field from point charges is $E=\frac{kq}{r^2}$ , where we know the values of the following variables.

$k=8.99* 10^9\ \frac{\text{Nm}^2}{\text{C}^2}$

$q=100\ \text{nC}= 100*10^{-9}\ \text{C}$

$r=500\ \mu\text{m}= 500*10^{-6}\ \text{m}$

Using these values, we can solve for the electric field.

$E=\frac{(8.99* 10^9\ \frac{\text{Nm}^2}{\text{C}^2})(100*10^{-9}\ \text{C})}{(500*10^{-6}\ \text{m})^2}$

$E=3.6* 10^9 \frac{\text{N}}{\text{C}}$

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Example Question #2 : Using Coulomb's Law

Two positive point charges of q_1 and q_2 are place at a distance $1000\mu m$ away from each other, as shown below. If a positive test charge, , is placed in between, at what distance away from q_1 will this test charge experience zero net force?

Charges

Possible Answers:

$9.7*10^{-2}m$

$4.3*10^{-4}m$

$5.1*10^{-3}m$

$4.5*10^{-4}m$

$4.9*10^{-3}m$

Correct answer:

$4.5*10^{-4}m$

Explanation:

To find the location at which the test charge experience zero net force, write the net force equation as $F_{net}=F_1-F_2=0$ , where F_1 is the force on the test charge from q_1 , and F_2 is the force on the same test charge from q_2 . Using Coulomb's law, we can rewrite the force equation and set it equal to zero.

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

$F_{net}=\frac{kq_1q}{r^2}-\frac{kq_2q}{(d-r)^2}=0$

In this equation, the distance, , is how far away the test charge is from q_1 , while (d-r) represents how far away the test charge is from q_2 . Now, we simplify and solve for .

Cross-multiply.

kq_1q(d-r)^2=kq_2qr^2

We can cancel and . We do not need to know these values in order to solve the question.

q_1(d-r)^2=q_2r^2

$\sqrt{q_1}(d-r)=\sqrt{q_2}r$

$\sqrt{q_1}d-\sqrt{q_1}r=\sqrt{q_2}r$

$\sqrt{q_1}d=r(\sqrt{q_2}+\sqrt{q_1})$

$r=\frac{\sqrt{q_1}d}{\sqrt{q_2}+\sqrt{q_1}}$

Now that we have isolated , we can plug in the values given in the question and solve.

$r=\frac{\sqrt{21* 10^{-9}\ \text{C}}(1000*10^{-6}\ \text{m})}{\sqrt{32*10^{-9}\ \text{C}}+\sqrt{21* 10^{-9}\ \text{C}}}$

$r=\frac{(1.45*10^{-4}\ \text{C})(1000*10^{-6}\ \text{m})}{1.79*10^{-4}\ \text{C}+1.45*10^{-4}\ \text{C}}$

$r=4.5*10^{-4}\ \text{m}$

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Example Question #3 : Electricity

You are standing on top of a very large positively charged metal plate with a surface charge of $\sigma=9\times10^{-8}\ \frac{\text{C}}{\text{m}^2}$ .

Assuming that the plate is infinitely large and your mass is 67kg , how much charge does your body need to have in order for you to float?

$\epsilon_0=8.85*10^{-12}\frac{C^2}{Nm^2}$

Possible Answers:

3.1C

0.13mC

2.5C

0.13C

$5.8\mu C$

Correct answer:

0.13C

Explanation:

Consider the forces that are acting on you. There is the downward (negative direction) force of gravity, . In order for you to float, there has to be an upward (positive direction) force, and that upward force is coming from the metal plate, . To show that you would float, the net force equation is written as $F_{net}=qE-mg=0$ , where is the charge on you.

For plates that are charged, know that $E=\frac{\sigma}{2\epsilon_0}$ .

Knowing this, the force equation becomes $q\frac{\sigma}{2\epsilon_0}-mg=0$ .

Solve for .

$q=\frac{2\epsilon_0mg}{\sigma}$

Now we can plug in our given values, and solve for the charge.

$q=\frac{2(8.85*10^{-12}\ \frac{\text{C}^2}{\text{Nm}^2})(67\ \text{kg})(9.8\ \frac{\text{m}}{\text{s}^2})}{9*10^{-8}\ \frac{\text{C}}{\text{m}^2}}$

$q=\frac{1.2*10^{-8}\frac{C^2}{m^2}}{9*10^{-8}\frac{C}{m^2}}$

$q=0.13\ \text{C}$

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Example Question #1 : Using Coulomb's Law

A point charge of $q_1=10{nC}$ exerts a force of 235N on another charge with $q_2=2{nC}$ . How far apart are the two charges?

$k=8.99*10^9\frac{\text{Nm}^2}{\text{C}^2}$

Possible Answers:

$4.9*10^{4}m$

$1.2*10^{-2}m$

$5.82*10^{-4}m$

$7.5*10^{-6}m$

$2.77*10^{-5}m$

Correct answer:

$2.77*10^{-5}m$

Explanation:

To find the distance between the two charges, use Coulomb's Law.

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

Since we want to find distance, , we solve for .

$r=\sqrt{\frac{kq_1q_2}{F}}$

We know the values of the force and the two charges.

$k=8.99*10^9\frac{\text{Nm}^2}{\text{C}^2}$

$q_1=10{nC} = 10* 10^{-9}{C}$

$q_2=2{nC}=2* 10^{-9}{C}$

$F=235\ \text{N}$

We can plug in these values and solve for the distance.

$r=\sqrt{\frac{(8.99*10^9\frac{{Nm}^2}{{C}^2})(10* 10^{-9}{C})(2* 10^{-9}{C})}{235{N}}}$

$r=\sqrt{\frac{1.8*10^{-7}N*m^2}{235{N}}}$

$r=\sqrt{7.65*10^{-10}m^2}$

$r=2.77* 10^{-5}{m}$

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Example Question #2 : Using Coulomb's Law

What is the electric force between two charges, q_1=50nC and q_2=50nC , located 5cm apart?

$\epsilon_0=8.85* 10^{-12} \frac{F}{m}$

Possible Answers:

$4.5*10^{-3}N$

$1.8*10^{-3}N$

$6.7*10^{-3}N$

$9.0*10^{-3}N$

Correct answer:

$9.0*10^{-3}N$

Explanation:

The equation for finding the electric force between two charges is $F=\frac{kq_1q_2}{r^2}$ , where $k=\frac{1}{4\pi \epsilon_0}$ . Using this, we can rewrite the force equation.

$F=\frac{1}{4\pi \epsilon_0}\frac{q_1q_2}{r^2}$

Now, we can use the values given in the question to solve for the electric force between the two particles.

$F=\frac{1}{4\pi(8.85*10^{-12}\frac{F}{m})}\frac{(50*10^{-9}C)(50*10^{-9}C)}{(0.05m)^2}$

$F=\frac{1}{1.11*10^{-10}\frac{F}{m}}\frac{2.5*10^{-15}C^2}{0.0025m^2}$

$F=9*10^{9}\frac{m}{F}(1*10^{-12}\frac{C^2}{m^2})$

$F=9*10^{-3}N$

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Example Question #1 : Using Coulomb's Law

What is the magnitude of the electric field at a field point from a point charge of q=5nC ?

$\epsilon_0=8.85* 10^{-12} \frac{F}{m}$

Possible Answers:

$2.25\frac{N}{C}$

$180.0\frac{N}{C}$

$45.0\frac{N}{C}$

$11.25\frac{N}{C}$

Correct answer:

$45.0\frac{N}{C}$

Explanation:

The equation to find the strength of an electric field is $E=\frac{kq}{r^2}=\frac{1}{4\pi \epsilon_0}\frac{q}{r^2}$ .

We can use the given values to solve for the strength of the field at a distance of .

$E=\frac{1}{4\pi(8.85*10^{-12}\frac{F}{m})}\frac{5*10^{-9}C}{1m}$

$E=\frac{5*10^{-9}C}{1.11*10^{-10}\frac{F}{m}}$

$E=45\frac{N}{C}$

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Example Question #5 : Using Coulomb's Law

Two capacitors are in parallel, with capacitance values of 6mF and 10mF . What is their equivalent capacitance?

Possible Answers:

16mF

12mF

2.67mF

4mF

Correct answer:

16mF

Explanation:

The equivalent capacitance for capacitors in parallel is the sum of the individual capacitance values.

$C_{eq}=C_1+C_2$

Using the values given in the question, we can find the equivalent capacitance.

$C_{eq}=10mF+6mF=16mF$

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Example Question #11 : Electricity And Magnetism Exam

A proton moves in a straight line for a distance of . Along this path, the electric field is uniform with a value of $2*10^7 \frac{V}{m}$ . Find the force on the proton.

The charge of a proton is $1.61*10^{-19}\text{C}$ .

Possible Answers:

$3.22*10^{-12}N$

$1.61*10^{-12}N$

$6.44*10^{-12}N$

$0.8*10^{-12}N$

Correct answer:

$3.22*10^{-12}N$

Explanation:

The force of an electric field is given by the equation F=qE , where is the charge of the particle and is the electric field strength. We can use the given values from the question to solve for the force.

$F=qE=(1.61*10^{-19}C)(2*10^{7}\frac{V}{m})=3.22*10^{-12}N$

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Example Question #3 : Using Coulomb's Law

Two point charges, Q_1 and Q_2 are separated by a distance .

Ps0_twochargeefield

The values of the charges are:

$Q_{1} = - \,2\;nC$

$Q_{2} = + \,5\;nC$

The distance is 4.0cm. The point lies 1.5cm away from Q_1 on a line connecting the centers of the two charges.

What is the magnitude and direction of the net electric field at point due to the two charges?

$k=9*10^9\frac{Nm^2}{C^2}$

Possible Answers:

$8000\ \frac{V}{m},\ \text{toward Q}_1$

$152000\ \frac{V}{m},\ \text{toward Q}_2$

$8000\ \frac{V}{m},\ \text{toward Q}_2$

Correct answer:

Explanation:

At point , the electric field due to Q_1 points toward Q_1 with a magnitude given by:

$E_{1}=\left| \frac{kQ_{1}}{r_{1}^{2}} \right| =\left| \frac{(9*10^9\frac{Nm^2}{C^2})(-2nC)}{(1.5cm)^{2}} \right| =80000 \;\frac{V}{m}$

At point P, the electric field due to Q₂ points away from Q₂ with a magnitude given by

$E_{2}=\left| \frac{kQ_{2}}{r_{2}^{2}} \right|= \left| \frac{(9*10^9\frac{Nm^2}{C^2})(5nC)}{(4.0-1.5cm)^{2}} \right| =72000 \;\frac{V}{m}$

The addition of these two vectors, both pointing in the same direction, results in a net electric field vector of magnitude 152000 volts per meter, pointing toward Q_1 .

$80000\frac{V}{m}+72000\frac{V}{m}=152000\frac{V}{m}$

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All AP Physics C Electricity Resources

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AP Physics C Electricity : Using Coulomb's Law

Study concepts, example questions & explanations for AP Physics C Electricity

All AP Physics C Electricity Resources

Example Questions

Example Question #1 : Electricity And Magnetism Exam

Example Question #2 : Using Coulomb's Law

Example Question #3 : Electricity

Example Question #1 : Using Coulomb's Law

Example Question #2 : Using Coulomb's Law

Example Question #1 : Using Coulomb's Law

Example Question #5 : Using Coulomb's Law

Example Question #11 : Electricity And Magnetism Exam

Example Question #3 : Using Coulomb's Law

All AP Physics C Electricity Resources

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