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 cahrges compare to the original force

it will decrease to 6 times the original force
it willnincrease to 36 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

I just had this question on my test, the answer is C.

Answer 2

The new force will decrease to one sixth the original value.

Step-by-step explanation:

It is given that,

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

Let,
`q'_1=(1)/(3)q_1` q₁ is the charge 1

`q'_2=(1)/(2)q_2` q₂ is the charge 2

Initially, the electrical force between the charges is given by :

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

Let F' is the new force i.e.

`F'=k(q'_1q'_2)/(r^2)`

So,

`F'=k((1)/(3)q_1* (1)/(2)q_2)/(r^2)`

`F'=(1)/(6)k(q_1q_2)/(r^2)`

`F'=(1)/(6)F` (from equation 1)

Hence, the new force will decrease to one sixth the original value.