Question

One charge is decreased to one-third of its original value,

and a second charge is decreased to one-half of its
original value.
How will the electrical force between the charges compare
with the original force?
o It will increase to six times the original force.
O It will increase to thirty-six times the original force.
It will decrease to one-sixth the original force.
It will decrease to one-thirty-sixth the original force.

Answer 1

the Third option

Step-by-step explanation:

Answer 2

It will decrease to one-sixth the original force.

Step-by-step explanation:

The electrical force is given by :

`F=k(q_1q_2)/(r^2)`..............(1)

Where

k is the electrostatic constant

r is the distance between charges

One charge is decreased to one-third of its original value, and a second charge is decreased to one-half of its original value.

i.e.
`q_1'=(q_1)/(3)`

and
`q_2'=(q_2)/(2)`

New force is given by :

`F'=k(q_1q_2)/(6r^2)`

`F'=(F)/(6)`

So, the new force will decrease to one-sixth the original force. Hence, the correct option is (c).