Question

how many newtons of force will 2 pieces of paper have if one has a charge of .2 Coilombs and the other has a charge of .3 Coulombs and they are .005 meters apart?

What will happen if they are the same charge (+ or - ) Opposite charges ?

Answer 1

1) The electrostatic force between the two pieces of paper is $2.16\cdot 10^{13} N$

2)

• If the charges of the two pieces of paper are both + or both -, they repel each other
• If the charges of one piece of paper is positive and the other one is negative, they attract each other

Explanation:

1)

The electrostatic force between two charged objects is given by Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the two charges

r is the separation between the two charges

In this problem, we have the following situation:

$q_1 = 0.2 C$ is the charge on the first piece of paper

$q_2 = 0.3 C$ is the charge on the second piece of paper

$r=0.005 m$ is their separation

Substituting into the equation, we find the magnitude of the electrostatic force between them:

$F=k\frac{q_1 q_2}{r^2}=(8.99\cdot 10^9) \frac{(0.2)(0.3)}{(0.005)^2}=2.16\cdot 10^{13} N$

2)

The electrostatic force can be either attractive of repulsive, depending on the relative sign of the charges of the objects involved.

In particular, we have:

• If the two charges have same sign (both positive or both negative), the force between them is repulsive
• If the two charges have opposite sign (one positive and one negative), the force between them is attractive

Therefore, in this case:

• If the charges of the two pieces of paper are both + or both -, they repel each other
• If the charges of one piece of paper is positive and the other one is negative, they attract each other

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