Question

two point charges of +2 micro coulomb and +6 microcoulomb repel each other with a force of 12N.if each is given an additional charge of -4 microcoulomb, what will be the new force

Answer 1

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Answer 2

Answer:  

The new force is attractive in nature as the charges are in opposite sign and the magnitude of the new force is $\bold{4 \times 10^{-6} N}$.

Solution:

The old charges are +2μC and +6μC. If an additional charge of -4μC is added in that then the new charges will be -2μC and 2μC respectively.

The force between the two old point charges is 12N. As they both are positive charges, they repel each other. So the distance between the two charges can be found from coulomb’s law.

$F=k \frac{q q^{\prime}}{r^{2}}$

$12=9 \times 10^{9} \times \frac{2 \times 10^{-6} \times 6 \times 10^{-6}}{r^{2}}$

$r^{2}=\frac{9 \times 2 \times 6 \times 10^{3}}{12}=9 \times 10^{3}$

The distance found can be used in the force between the new charges -2μC and 2μC.

$F=9 \times 10^{9} \times \frac{2 \times 10^{-6} \times-2 \times 10^{-6}}{9 \times 10^{3}}$

$F=-\frac{9 \times 2 \times 2 \times 10^{9-6-6-3}}{9}=-4 \times 10^{-6} N$

So, the new force is attractive in nature as the charges are in opposite sign and the magnitude of the new force is $\bold{4 \times 10^{-6} \mathrm{N}.}$