Question

# A charge of 6.00\times10^{-9} C and a charge of -3.00\times10^{-9} C are separated by a distance of 60.0 cm. Find the position at which a third charge, of 12.0\times10^{-9} C, can be placed so that the net electrostatic force on it is zero.

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A charge of $$\displaystyle{6.00}\times{10}^{{-{9}}}$$ C and a charge of $$\displaystyle-{3.00}\times{10}^{{-{9}}}$$ C are separated by a distance of 60.0 cm. Find the position at which a third charge, of $$\displaystyle{12.0}\times{10}^{{-{9}}}$$ C, can be placed so that the net electrostatic force on it is zero.

2021-05-05
Let $$\displaystyle{q}_{{1}}={6.00}\times{10}^{{-{9}}}$$ C
$$\displaystyle{q}_{{2}}={3.00}\times{10}^{{-{9}}}$$ C
The distance of seperation =60 cm
$$\displaystyle={0.6}{m}$$
Let the position of the third charge $$\displaystyle{q}_{{3}}$$ beat a distance (x) from the charge $$\displaystyle{q}_{{2}}$$ now as the net electrostatic force on the charge $$\displaystyle{q}_{{3}}$$ must be zero, the forces due to $$\displaystyle{q}_{{1}}$$ and $$\displaystyle{q}_{{2}}$$ on $$\displaystyle{q}_{{3}}$$ must be equal in magnitude and oppsite, so that they will canceal each other and the netelectrostatic force is zero
then the force $$\displaystyle{F}_{{1}}$$ on $$\displaystyle{q}_{{3}}$$ due to $$\displaystyle{q}_{{1}}$$ will be
$$\displaystyle{F}_{{1}}={\frac{{{k}{\left({q}_{{1}}{q}_{{3}}\right)}}}{{{\left({0.6}+{x}\right)}^{{2}}}}}$$
and the force $$\displaystyle{F}_{{2}}$$ on $$\displaystyle{q}_{{3}}$$ due to $$\displaystyle{q}_{{2}}$$ will be
$$\displaystyle{F}_{{1}}={\frac{{{k}{\left({q}_{{2}}{q}_{{3}}\right)}}}{{{\left({x}\right)}^{{2}}}}}$$
as $$\displaystyle{F}_{{1}}={F}_{{2}}$$
Solve for x