Question

74.
Two point charges A and B, having charges
+Q and -Q respectively, are placed at certain
distance apart and force acting between them is
F. If 25% charge of A is transferred to B, then
force between the charges becomes :
Se le e
​

Answer 1

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Answer 2

Answer:

The force between the charges become $\dfrac{9F}{16}$

Explanation:

Given:

Magnitude of the charges are Q, -Q

The force between the charges=F

Let the radial distances between the charges  be r.

According to the coulombs law the force between the particle is equal to

$F=\dfrac{kQ\times Q}{r^2}\\F=\dfrac{kQ^2}{r^2}$

Now when the 25 percent of charge of A is transferred to B then

New magnitude of charge on A=$\dfrac{3Q}{4}$

charge on B=$\dfrac{-3Q}{4}$

So let $F_1$ be the magnitude of new force between them which is given by

$F_1=\dfrac{k\times \dfrac{3Q}{4}\times \dfrac{3Q}{4}}{r^2}\\F_1=\dfrac{9kQ^2}{16}\\F_1=\dfrac{9F}{16}$

Hence the new force is calculated.