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

Answer:

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

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, $q'_1=\dfrac{1}{3}q_1$ q₁ is the charge 1

$q'_2=\dfrac{1}{2}q_2$ q₂ is the charge 2

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

$F=k\dfrac{q_1q_2}{r^2}$...........(1)

Let F' is the new force i.e.

$F'=k\dfrac{q'_1q'_2}{r^2}$

So,

$F'=k\dfrac{\dfrac{1}{3}q_1\times \dfrac{1}{2}q_2}{r^2}$

$F'=\dfrac{1}{6}k\dfrac{q_1q_2}{r^2}$

$F'=\dfrac{1}{6}F$  (from equation 1)

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