Question

Two like-charged particles are placed close to each other. How would the force of repulsion be affected if the charge on one of the particles is doubled and that on the other is reduced to half the original value?

Answer 1

The Answer Is A my evidence is the formula above

Step-by-step explanation:

Answer 2

Answer: It will remain the same

Step-by-step explanation:

According to Coulomb's Law:

"The electrostatic force
`F_(E)` between two point charges
`q_(1)` and
`q_(2)` is proportional to the product of the charges and inversely proportional to the square of the distance
`d` that separates them, and has the direction of the line that joins them".

Mathematically this law is written as:

`F_(E)=K(q_(1).q_(2))/(d^(2))` (1)

Where:

`F_(E)` is the electrostatic force

`K` is the Coulomb's constant

`q_(1)=q_(2)=q` are the electric charges , which in this case have the same positive charge

`d` is the separation distance between the charges

Rewritting we have:

`F_(E)=K(q^(2))/(d^(2))` (2)

Now, if the first charge is doubled:

`q_(1)=2q_`

And the second is reduced to a half:

`q_(2)=(1)/(2)q`

We will have the following:

`F_(E)=K((2q)((1)/(2)q))/(d^(2))` (3)

`F_(E)=K(q^(2))/(d^(2))` (4)

As we can see equation (4) is equal to equation (2), this means the force of repulsion between both charges will remain the same