Using Coulomb's Law - AP Physics C: Electricity and Magnetism

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Question

Two point charges, and are separated by a distance .

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 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?

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

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Answer

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

$E_{1}=\left| \frac{kQ_{1}}{r_{1}^{2}} \right| =\left| \frac{(910^9\frac{Nm^2}{C^2})(-2nC)}{(1.5cm)^{2}} \right| =80000 ;\frac{V}{m}$

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 .

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

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Question

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 , how much charge does your body need to have in order for you to float?

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

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Answer

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.

$q=\frac{2(8.8510^{-12}\ \frac{\text{C}^2}{\text{Nm}^2})(67\ \text{kg})(9.8\ \frac{\text{m}}{\text{s}^2})}{910^{-8}\ \frac{\text{C}}{\text{m}^2}}$

$q=\frac{1.210^{-8}\frac{C^2}{m^2}}{910^{-8}\frac{C}{m^2}}$

$q=0.13\ \text{C}$

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Question

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

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

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Answer

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.

$k=8.9910^9\frac{\text{Nm}^2}{\text{C}^2}$

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

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

$F=235\ \text{N}$

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

$r=\sqrt{\frac{(8.9910^9\frac{{Nm}^2}{{C}^2})(10 10^{-9}{C})(2* 10^{-9}{C})}{235{N}}}$

$r=\sqrt{\frac{1.810^{-7}Nm^2}{235{N}}}$

$r=\sqrt{7.6510^{-10}m^2}$

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

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Question

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

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

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Answer

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.

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

$q=100\ \text{nC}= 10010^{-9}\ \text{C}$

$r=500\ \mu\text{m}= 50010^{-6}\ \text{m}$

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

$E=\frac{(8.99* 10^9\ \frac{\text{Nm}^2}{\text{C}^2})(10010^{-9}\ \text{C})}{(50010^{-6}\ \text{m})^2}$

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

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Question

Two positive point charges of and 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 will this test charge experience zero net force?

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Answer

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 is the force on the test charge from , and is the force on the same test charge from . 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 , while represents how far away the test charge is from . Now, we simplify and solve for .

Cross-multiply.

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

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

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

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

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

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

$r=\frac{\sqrt{21* 10^{-9}\ \text{C}}(100010^{-6}\ \text{m})}{\sqrt{3210^{-9}\ \text{C}}+\sqrt{21* 10^{-9}\ \text{C}}}$

$r=\frac{(1.4510^{-4}\ \text{C})(100010^{-6}\ \text{m})}{1.7910^{-4}\ \text{C}+1.4510^{-4}\ \text{C}}$

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

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Question

What is the electric force between two charges, and , located apart?

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

Tap to reveal answer

Answer

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.

$F=\frac{1}{4\pi(8.8510^{-12}\frac{F}{m})}\frac{(5010^{-9}C)(5010^{-9}C)}{(0.05m)^2}$

$F=\frac{1}{1.1110^{-10}\frac{F}{m}}\frac{2.510^{-15}C^2}{0.0025m^2}$

$F=910^{9}\frac{m}{F}(110^{-12}\frac{C^2}{m^2})$

$F=910^{-3}N$

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Question

What is the magnitude of the electric field at a field point from a point charge of ?

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

Tap to reveal answer

Answer

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 .

$E=\frac{1}{4\pi(8.8510^{-12}\frac{F}{m})}\frac{510^{-9}C}{1m}$

$E=\frac{510^{-9}C}{1.1110^{-10}\frac{F}{m}}$

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

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Question

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

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Answer

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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Question

A proton moves in a straight line for a distance of . Along this path, the electric field is uniform with a value of $210^7 \frac{V}{m}$ . Find the force on the proton.

The charge of a proton is $1.6110^{-19}\text{C}$ .

Tap to reveal answer

Answer

The force of an electric field is given by the equation , 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.

$F=qE=(1.6110^{-19}C)(210^{7}\frac{V}{m})=3.22*10^{-12}N$

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Question

Two point charges, and are separated by a distance .

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 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?

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

Tap to reveal answer

Answer

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

$E_{1}=\left| \frac{kQ_{1}}{r_{1}^{2}} \right| =\left| \frac{(910^9\frac{Nm^2}{C^2})(-2nC)}{(1.5cm)^{2}} \right| =80000 ;\frac{V}{m}$

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 .

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

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Question

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 , how much charge does your body need to have in order for you to float?

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

Tap to reveal answer

Answer

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.

$q=\frac{2(8.8510^{-12}\ \frac{\text{C}^2}{\text{Nm}^2})(67\ \text{kg})(9.8\ \frac{\text{m}}{\text{s}^2})}{910^{-8}\ \frac{\text{C}}{\text{m}^2}}$

$q=\frac{1.210^{-8}\frac{C^2}{m^2}}{910^{-8}\frac{C}{m^2}}$

$q=0.13\ \text{C}$

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Question

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

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

Tap to reveal answer

Answer

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.

$k=8.9910^9\frac{\text{Nm}^2}{\text{C}^2}$

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

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

$F=235\ \text{N}$

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

$r=\sqrt{\frac{(8.9910^9\frac{{Nm}^2}{{C}^2})(10 10^{-9}{C})(2* 10^{-9}{C})}{235{N}}}$

$r=\sqrt{\frac{1.810^{-7}Nm^2}{235{N}}}$

$r=\sqrt{7.6510^{-10}m^2}$

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

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Question

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

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

Tap to reveal answer

Answer

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.

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

$q=100\ \text{nC}= 10010^{-9}\ \text{C}$

$r=500\ \mu\text{m}= 50010^{-6}\ \text{m}$

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

$E=\frac{(8.99* 10^9\ \frac{\text{Nm}^2}{\text{C}^2})(10010^{-9}\ \text{C})}{(50010^{-6}\ \text{m})^2}$

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

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Question

Two positive point charges of and 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 will this test charge experience zero net force?

Tap to reveal answer

Answer

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 is the force on the test charge from , and is the force on the same test charge from . 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 , while represents how far away the test charge is from . Now, we simplify and solve for .

Cross-multiply.

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

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

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

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

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

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

$r=\frac{\sqrt{21* 10^{-9}\ \text{C}}(100010^{-6}\ \text{m})}{\sqrt{3210^{-9}\ \text{C}}+\sqrt{21* 10^{-9}\ \text{C}}}$

$r=\frac{(1.4510^{-4}\ \text{C})(100010^{-6}\ \text{m})}{1.7910^{-4}\ \text{C}+1.4510^{-4}\ \text{C}}$

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

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Question

What is the electric force between two charges, and , located apart?

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

Tap to reveal answer

Answer

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.

$F=\frac{1}{4\pi(8.8510^{-12}\frac{F}{m})}\frac{(5010^{-9}C)(5010^{-9}C)}{(0.05m)^2}$

$F=\frac{1}{1.1110^{-10}\frac{F}{m}}\frac{2.510^{-15}C^2}{0.0025m^2}$

$F=910^{9}\frac{m}{F}(110^{-12}\frac{C^2}{m^2})$

$F=910^{-3}N$

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Question

What is the magnitude of the electric field at a field point from a point charge of ?

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

Tap to reveal answer

Answer

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 .

$E=\frac{1}{4\pi(8.8510^{-12}\frac{F}{m})}\frac{510^{-9}C}{1m}$

$E=\frac{510^{-9}C}{1.1110^{-10}\frac{F}{m}}$

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

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Question

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

Tap to reveal answer

Answer

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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Question

A proton moves in a straight line for a distance of . Along this path, the electric field is uniform with a value of $210^7 \frac{V}{m}$ . Find the force on the proton.

The charge of a proton is $1.6110^{-19}\text{C}$ .

Tap to reveal answer

Answer

The force of an electric field is given by the equation , 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.

$F=qE=(1.6110^{-19}C)(210^{7}\frac{V}{m})=3.22*10^{-12}N$

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Question

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

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

Tap to reveal answer

Answer

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.

$k=8.9910^9\frac{\text{Nm}^2}{\text{C}^2}$

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

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

$F=235\ \text{N}$

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

$r=\sqrt{\frac{(8.9910^9\frac{{Nm}^2}{{C}^2})(10 10^{-9}{C})(2* 10^{-9}{C})}{235{N}}}$

$r=\sqrt{\frac{1.810^{-7}Nm^2}{235{N}}}$

$r=\sqrt{7.6510^{-10}m^2}$

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

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Question

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

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

Tap to reveal answer

Answer

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.

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

$q=100\ \text{nC}= 10010^{-9}\ \text{C}$

$r=500\ \mu\text{m}= 50010^{-6}\ \text{m}$

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

$E=\frac{(8.99* 10^9\ \frac{\text{Nm}^2}{\text{C}^2})(10010^{-9}\ \text{C})}{(50010^{-6}\ \text{m})^2}$

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

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