Question

Two charged particles attract each other with a force of magnitude F. If the distance between the charges is made 3.5 times as large and the charge on one of the particles is made 4 times as big, what is the ratio of the new F to the old one

Answer 1

The ratio between the forces is:

`(F_(new))/(F_(old))=0.33`

Step-by-step explanation:

The electrostatic force equation is:

`F_(old)=k(q_(1)q_(2))/(d^(2))`

Where:

q1 and q2 are the electric charges

d is the distance between them

k is the electrostatic constant

Now, the distance is 3.5 times as large and q1 is 4 times as big, then the new force will be:

`F_(new)=k(4q_(1)q_(2))/((3.5d)^(2))`

`F_(new)=(4)/(3.5^(2))k(q_(1)q_(2))/(d^(2))`

We can rewrite this equation in terms of F(old)

`F_(new)=(4)/(3.5^(2))F_(old)`

Therefore, the ratio between the forces is:

`(F_(new))/(F_(old))=(4)/(3.5^(2))`

`(F_(new))/(F_(old))=0.33`

I hope it helps you!