Question

Two-point charge $+3\mu Cand+8\mu C$ each other with a force of $40N$. If a charge of $-5\mu C$ is added to each of them,then the force between them will become:-

• A. $-10N$
• B. $+10N$
• C. $+20N$
• D. $-20N$

Answer 1

The correct option is B $+10N$

Two point charge $=+3\mu C$ and $+8\mu C$ force $=40N$

Now, $q_1=3+(-5)=-2$

$q_2=8+(-5)=3$

Case -1

$F=K\frac{q_1{q}_2}{r^2}$

or, $40=9\times {10}^9\times \frac{3\times 8\times {10}^{-12}}{r^2}$

or, $r^2=\frac{9\times {10}^9\times 24\times {10}^{-12}}{40}$

$=\frac{9\times 24}{40}\times {10}^{-3}$

$=\frac{9\times 24}{4}\times {10}^{-4}$

$=54\times {10}^{-4}$

or, $r=\sqrt{54}\times {10}^{-2}m$

$=0.073m$

now,

Case-2

$F=9\times {10}^9\times \frac{6\times {10}^{-12}}{{\left(0.073\right)}^2}$

$=10.13N$

$\approx 10N$