Question

Coulomb's Law

Master Page: https://questioncove.com/study#/updates/5ae2c3a7293a9d6f092ba200

Answer 1

Coulomb's Law Formula:

\[F = \frac{ k q _{1} q _{2}}{ r^2 }\]

Where,
F = the electrical force between two charges (in Newtons)
q1 and q1 = the charges (in Coulombs)
r = the distance between the charges (in meters)
k = the constant (known as Coulomb's constant
\[k = 9 \times 10^{9}\]

Keep in mind that when the force is negative, q1 and q2 are opposite charges to each other, and thus the force between them is attractive.

\[F = \frac{ k (1)(-1) }{ r^2 } = -F\]

If the force is positive, then the charges are either both positive or both negative, which results in a repulsive force between the charges.

The test will likely incorporate the following:
\[\mu = 10^{-6}\]
\[\eta = 10^{-9}\]

Know the values, as they were on the last quiz.

Now onto the study guide...

Answer 2

Quick note about the relationships in regards to the formula derived from Coulomb's Law.

\[F = \frac{ k q _{1} q _{2} }{ r ^2}\]

The electrical force between two charges is directly proportional to the strength of each charge.

The electrical force between two charges is inversely proportional to the square of the distance between the charges.

Example:

Increasing the strength of q1 by two, and the strength of q2 by four will increased the electrical force by a factor of eight. This is because:

\[F = \frac{ k (2q _{1})(4q _{2}) }{ r ^2}= \frac{ k \times 8q _{1} q_{2}}{ r^2}\]

Increasing the distance by a factor of three will decrease the electrical force between the two charges by a factor of nine.

The key is to remember that it's inversely proportional to the 'square' of the distance. Take the factor by which the distance was increased, square it, then that's how much the electrical force has decreased by.

Answer 3

Coulomb's Law - Practice 1

1. What happens to the force between two charges if the distance between them is tripled?

It decreases by a factor of 9

2. What happens to the force between two charges if the distance between them is quadrupled?

It decreases by a factor of 16

3. What happens to the force between two charges if the distance between them is cut by half?

It increases by a factor of 2

4. What happens to the force between two charges if the magnitude of one charge is doubled?

It increases by a factor of 2

5. What happens to the force between two charges if the magnitude of both charges is doubled?

It increases by a factor of 4

6. What happens to the force between two charges if the magnitude of both charges is doubled and the distance between them is doubled.

Nothing
\[(2 \times 2) \times (2^{2})^{-1} = 4 \times 4^{-1} = 4 \times \frac{ 1 }{ 4 } = 1\]
Remember, square the distance, then take the inverse

7. What happens to the force between two charges if the magnitude of both charges is doubled and the distance between them is cut in half?

It increases by a factor of 16
\[(2 \times 2) \times ((\frac{ 1 }{ 2 })^{2})^{-1} = 4 \times (\frac{ 1 }{ 4 })^{-1} = 4 \times 4 = 16\]

Answer 4

Coulomb's Law - Practice 2
\[F = \frac{ k q _{1}q_{2} }{ r^2}\]
1. Two particles, each with a charge of 1 C, are separated by a distance of 1 meter. What is the force between the particles?

\[F = \frac{ (9 \times 10^{9})(1)(1) }{(1)^2}\]
\[F = \frac{ 9 \times 10^9 }{ 1} = 9 \times 10^9 N\]

2. What is the force between a 3 C charge and a 2 C charge separated by a distance of 5 meters?

\[F = \frac{ (9 \times 10^{9})(3)(2) }{(5)^2}\]
\[F = \frac{ 5.4 \times 10^{10} }{ 25 } = 2.16 \times 10^9 N \]

3. Calculate the force between a 0.006 C charge and a 0.001 C charge 4 meters apart.

\[F = \frac{ (9 \times 10^{9}(0.006)(0.001) }{ (4)^2}\]
\[F = \frac{ 54,000 }{ 16 } = 3, 375 N\]

4. Calculate the force between a 0.05 C charge and a 0.03 C charge 2 meters apart.

\[F = \frac{ (9 \times 10^9(0.05)(0.03) }{ (2)^2 }\]
\[F = \frac{ 13, 500, 000 }{ 4 } = 3, 375, 000 N\]

5. Two particles are each given a charge of 5 x 10^-5 C. What is the force between the charged particles if the distance between them is 2 meters?

\[F = \frac{ (9 \times 10^{9})(5 \times 10^{-5})(5 \times 10^{-5}) }{ (2)^2 } = \frac{ (9 \times 10^{9})(5 \times 10^{-5})^{2} }{4}\]
\[F = \frac{ 22.5 }{ 4} = 5.625 = 5.63 N\]

6. The force between a pair of charges is 100 newtons. The distance between the charges is 0.01 meter. If one of the charges is 2 times 10^-10 C, what is the strength of the other charge?

\[F = \frac{ k q_{1} q_{2} }{ r^2 }\]
\[F \times r^2 = k q_{1}q_{2}\]
\[\frac{ F \times r^2 }{ k q_{1} } = q_{2}\]
\[\frac{ (100) \times (0.01)^2 }{ (9 \times 10^{9})(2 \times 10^{-10} } = q_{2}\]
\[\frac{ 0.01 }{ 1.8 } = q_{2}\]
\[q_{2} = 0.0055...= 0.006 C\]

7. Two equal charges separated by a distance of 1 meter experience a repulsive force of 1,000 newtons. What is the strength in coulombs of each charge?

Note that it is 'two equal charges' thus q1 = q2, which can be represented as q^2

\[F = \frac{ k q^2 }{ r^2 }\]
\[F \times r^2 = k q^{2}\]
\[\frac{ F \times r^2 }{ k } = q^2\]
\[q = \sqrt(\frac{ F \times r^2 }{ k })\]
\[q = \sqrt(\frac{ 1,000 \times 1^2 }{ 9 \times 10^9})\]
\[q = \sqrt (\frac{ 1 }{ 9, 000, 000})\]

\[q = \frac{ 1 }{ 3000 } = 0.0003... = \pm 0.0003 C\]

8. The force between a pair of 0.001 C charges is 200 N. What is the distance between them?

\[F = \frac{ k q_{1} q_{2} }{ r^2 }\]
\[F \times r^2 = k q_{1} q_{2}\]
\[r^2 = \frac{ k q_{1} q_{2} }{ F }\]
\[r = \sqrt( \frac{ k q_{1} q_{2} }{ F })\]
\[r = \sqrt( \frac{ k q_{1} q_{2} }{ F })\]
\[r = \sqrt (\frac{ k q^2 }{ F})\]

The last step recognizes the fact that both charges are 0.001

\[r = \sqrt( \frac{ (9 \times 10^9) (0.001)^2 }{ 200 })\]
\[r = \sqrt( \frac{ 9000 }{ 200 }) = 3 \sqrt 5 = 6.71 m\]

9. The force between two charges is 1000 N. One has a charge of 2 times 10^-5 C, and the other has a charge of 5 times 10^-6 C. What is the distance between them

Lets grab the second to last step from the above question when we were solving for r.

\[r = \sqrt( \frac{ k q_{1} q_{2} }{ F })\]
\[r = \sqrt( \frac{ (9 \times 10^9)(2 \times 10^{-5})(5 \times 10^{-6}) }{ 1000 })\]
\[r = \sqrt (\frac{ 0.9 }{ 1000 }) = \sqrt (0.0009) = 0.00095 m\]

10. The forces between two charges is 2 newtons. The distance between the charges is 2 times 10^-4 m. If one of the charges is 3 times 10^-6 C, what is the strength of the other charge?

\[\frac{ F \times r^2 }{ k q_{1} } = q_{2}\]

Refer to problem #6 as to where I got this formula from

\[q_{2} = \frac{ (2) \times (2 \times 10^{-4})^{2} }{ (9 \times 10^9 )(3 \times 10^{-6}) }\]
\[q_{2} = \frac{ 8 \times 10^{-8} }{ 27,000 }\]
\[q_{2} = 3.0 \times 10^{-12} C\]