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AP Physics 1 : Coulomb's Law

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Example Question #1 : Coulomb's Law

Two protons are on either side of an electron as shown below:

Diagram

The electron is 30 µm away from the proton on its left and 10 µm away from the proton on its right. What is the magnitude and direction of the net electric force acting on the electron?

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

A proton has a charge of $+1.60 \cdot 10^{-19}C$

Possible Answers:

$7.67\cdot 10^{-24} N$ to the right

$7.67\cdot 10^{-24} N$ to the left

$-9.59\cdot10^{-24}N$ to the right

$2.04 \cdot 10^{-18} N$ to the right

$2.04 \cdot 10^{-18}N$ to the left

Correct answer:

$2.04 \cdot 10^{-18} N$ to the right

Explanation:

The net force on the electron is the sum of the forces between the electron and each of the protons:

$F_{net}=F_1+F_2$

These forces are given by Coulomb's law:

$F=k\frac{q_1q_2}{r^2}$

Using the numbers given, we get:

$F_1=(8.99\cdot10^9)\frac{(+1.60 \cdot 10^{-19})(-1.60 \cdot 10^{-19})}{(30 \cdot 10^-6)^2} = 2.56 \cdot 10^{-19} N$

$F_2=(8.99\cdot10^9)\frac{(+1.60 \cdot 10^{-19})(-1.60 \cdot 10^{-19})}{(10 \cdot 10^-6)^2} = 2.30 \cdot 10^{-18} N$

Because opposite charges attract, F_1 points left (the negative direction) and F_2 points right (the positive direction).

Therefore, the net force is

$F_{net} = -2.56 \cdot 10^{-19} N + 2.30 \cdot 10^{-18} N = 2.04 \cdot 10^{-18}N$

Because this value is positive, the direction is rightward.

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Example Question #1 : Coulomb's Law

Charges A and B are placed a distance of 1.5m from one another. The charge of particle A is 0.003C whereas the charge of particle B is 0.006C . Charge B experiences an electrostatic force of F_b from charge A. Similarly, charge A experiences an electrostatic force of F_a from charge B.

What is the ratio of F_a to F_b ?

Possible Answers:

2:1

1:2

1:3

1:1

Correct answer:

1:1

Explanation:

This question is very simple if you realize that the force experienced by both charges is equal.

The definition of the two electrostatic forces are given by Coulomb's law:

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

In this question, we can rewrite this equation in terms of our given system.

$F_a = -k\frac{q_a q_b}{r^2}=-k\frac{(0.003C)(0.006C)}{(1.5m)^2}=-k(8*10^{-6}\frac{C^2}{m^2})$

$F_b = -k\frac{q_b q_a}{r^2}=-k\frac{(0.006C)(0.003C)}{(1.5m)^2}=-k(8*10^{-6}\frac{C^2}{m^2})$

It doesn’t matter if the charges of the two particles are different; both particles experience the same force because the charges of both particles are accounted for in the electrostatic force equation (Coulomb's law). This conclusion can also be made by considering Newton's third law: the force of the first particle on the second will be equal and opposite the force of the second particle on the first.

Since the forces are equal, their ratio will be $\small 1:1$ .

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Example Question #71 : Ap Physics B

An excess charge of 10 mC is put on an ideal neutral conducting sphere with radius 20 cm . What is the Coulomb force this excess charge exerts on a point charge of 5 mC that is 30 cm from the surface of the sphere?

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

Possible Answers:

0.9 N

5.6 N

10.1 N

1.8 N

2.5 N

Correct answer:

1.8 N

Explanation:

Two principal realizations help with solving this problem, both derived from Gauss’ law for electricity:

1) The excess charge on an ideal conducting sphere is uniformly distributed over its surface

2) A uniform shell of charge acts, in terms of electric force, as if all the charge were contained in a point charge at the sphere’s center

With these realizations, an application of Coulomb’s law answers the question. If $q_{2}$ is the point charge outside the sphere, then the force $F_{q2}$ on $q_{2}$ is:

$F_{q2} = \frac{kq_{sphere}q_{2}}{r^{2}}$

In this equation, is Coulomb’s constant, is the excess charge on the spherical conductor, and is total distance in meters of $q_{2}$ from the center of the conducting sphere.

r=0.20m+0.30m=0.50m

Using the given values in this equation, we can calculate the generated force:

$F_{q2}=\frac{(8.99*10^9\frac{m^2N}{C^2})(10*10^{-6}C)(5*10^{-6}C)}{(0.5m)^2}$

$F_{q2}=\frac{0.4495m^2N}{0.25m^2}=1.798N$

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Example Question #1 : Coulomb's Law

If the distance between two charged particles is doubled, the strength of the electric force between them will __________.

Possible Answers:

quadruple

be quartered

be halved

double

remain unchanged

Correct answer:

be quartered

Explanation:

Coulomb's law gives the relationship between the force of an electric field and the distance between two charges:

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

The strength of the force will be inversely proportional to the square of the distance between the charges.

When the distance between the charges is doubled, the total force will be divided by four (quartered).

$F_e=k\frac{q_1q_2}{(2r)^2}=k\frac{q_1q_2}{4r^2}=\frac{1}{4}(k\frac{q_1q_2}{r^2})$

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Example Question #1 : 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:

2.7N

6.3N

7.2N

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:

67.31N

52.86N

71.63N

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:

2.7N

0.027N

27N

270N

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 #6 : 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:

180N

45N

15N

90N

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 #7 : 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:

13.4814N

8.9876N

26.9628N

0.89876N

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 #8 : 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

40.44J

11.23J

20.22J

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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AP Physics 1 : Coulomb's Law

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #1 : Coulomb's Law

Example Question #1 : Coulomb's Law

Example Question #71 : Ap Physics B

Example Question #1 : Coulomb's Law

Example Question #1 : Coulomb's Law

Example Question #4 : Coulomb's Law

Example Question #5 : Coulomb's Law

Example Question #6 : Coulomb's Law

Example Question #7 : Coulomb's Law

Example Question #8 : Coulomb's Law

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