Question

Two charged objects originally felt 12N of attraction. One charge

is changed from 3C to 6C and their distance changes from 15cm

apart to 45cm apart. What is the new force of attraction?

Answer 1

The new force of attraction will be
`F_(new)=2.67\: N`

Step-by-step explanation:

The electrostatic force equation is given by:

`F=k(q_(1)q_(2))/(r^(2))`

Where:

• q(1) is the first charge, let's use this one to the 3 C
• q(2) is the second charge
• r is the initial distance (r = 0.15 m)

Then, the original force could express as:

`F_(original)=k(3q_(2))/(0.15^(2))=12`

This force is 12 N.

Now, we know that q(1) changes its value to 6 C and the new distance between the charges is 0.45 m, which means:

`F_(new)=k(6q_(2))/(0.45^(2))`

We can rewrite this force in terms of the original force.

`F_(new)=k(2\cdot 3q_(2))/((3\cdot 0.15)^(2))`

`F_(new)=(2)/(9)k(3q_(2))/(0.15^(2))`

`F_(new)=(2)/(9)F_(original)`

`F_(new)=(2)/(9)12`

Therefore, the new force of attraction will be
`F_(new)=2.67\: N`

I hope it helps you!