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

\(\displaystyle{q}_{{2}}={3.00}\times{10}^{{-{9}}}\) 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