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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asked 2021-05-03
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.

Expert Answers (1)

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
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