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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.

Respuesta :

Answer:

the Third option

Explanation:

Answer:

It will decrease to one-sixth the original force.

Explanation:

The electrical force is given by :

[tex]F=k\dfrac{q_1q_2}{r^2}[/tex]..............(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. [tex]q_1'=\dfrac{q_1}{3}[/tex]

and [tex]q_2'=\dfrac{q_2}{2}[/tex]

New force is given by :

[tex]F'=k\dfrac{q_1q_2}{6r^2}[/tex]

[tex]F'=\dfrac{F}{6}[/tex]

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

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