Question

Objects 1 and 2 attract each other with a electrostatic force

of 36.0 units. If the charge of Object 1 is halved AND the
charge of object 2 is tripled, then the new electrostatic force
will be units.

Answer 1

According to Coulomb's Law, the electrostatic force between two charged objects is linearly proportional to the product of the charges and inversely proportional to the square of the distance between them. If the charge of Object 1 is halved and the charge of Object 2 is tripled, the new electrostatic force can be calculated by determining the new value of the product of the charges. In this scenario, the new electrostatic force will be 54.0 units.

Step-by-step explanation:

The electrostatic force between two charged objects is described by Coulomb's Law. According to Coulomb's Law, the magnitude of the force is linearly proportional to the product of the charges and inversely proportional to the square of the distance between them.

In this scenario, if the charge of Object 1 is halved and the charge of Object 2 is tripled, the new electrostatic force can be determined by calculating the new value of the product of the charges and using the same distance between the objects.

Let's assume the initial electrostatic force is 36.0 units. If the charge of Object 1 is halved, the new force will be (1/2) times the initial force. If the charge of Object 2 is tripled, the new force will be 3 times the initial force. Therefore, the new electrostatic force will be (1/2) × 3 × 36.0 units, which simplifies to 54.0 units.

Answer 2

54.0 units

Step-by-step explanation:

The electrostatic force between two charged particles is given by the equation:

`F=k(q_1 q_2)/(r^2)`

where

k is the Coulomb constant

`q_1, q_2` are the charges of the two particles

r is the separation between the particles

The direction of the force is:

- Repulsive if the two charges have same sign (++ or --)

- Attractive if the two charges have opposite sign (+-)

In this problem, the initial force is

F = 36.0 units

when we have charges
`q_1,q_2` and distance
`r`.

Later:

- Charge of object 1 is halved, so

`q_1'=(q_1)/(2)`

- Charge of object 2 is tripled, so

`q_2'=3q_2`

So, the new electrostatic force will be:

`F'=(kq_1' q_2')/(r^2)=(k((q_1)/(2))(3q_2))/(r^2)=(3)/(2)((kq_1 q_2)/(r^2))=(3)/(2)F`

And using F = 36.0, we find

`F'=(3)/(2)(36.0)=54.0`