Two particles with oppositely signed charges are held a fixed distance apart. The charges are equal in magnitude and they exert a force on each other. Half of one of the charges is transferred to the other charge and the distance between them is unchanged. What happens to the force exerted on one charge by the other charge? Group of answer choices The force is one fourth as large The force is doubled The force is unchanged The force is half as large The force is four times as large

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aristeus aristeus · Answer 1 · 2019-09-25T13:50:50+02:00

Answer:

Force will become [tex]\frac{3}{4}[/tex] times.

Explanation:

We have given two charge of opposite sign that are held at a fixed distance

According to coulombs law force between two charges is given by

[tex]F_1=\frac{1}{4\pi \epsilon _0}\frac{q_1q_1}{r^2}[/tex]

let initially the charges are [tex]q_1=q\ and\ q_2=q[/tex]

So force [tex]F=\frac{Kq^2}{r^2}[/tex]

Now according to question half of the charge is transferred to second charge particle

So [tex]q_1=\frac{q}{2}\ and\ q_2=\frac{3q}{2}[/tex]

As the distance remain unchanged

So force [tex]F_2=\frac{K3q^2}{4r^2}[/tex]

So [tex]\frac{F_1}{F_2}=\frac{4}{3}[/tex]

[tex]F_2=\frac{3}{4}F_1[/tex]

So force will become [tex]\frac{3}{4}[/tex] times.