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A charge q is placed at the centre of th...


A charge q is placed at the centre of the line joining two equal charges Q. Show that the 4 system of three charges will be in equilibrium if `q=-Q/4`.

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Let a charge q be placed at point such that its distance from charge at A = x and from charge at B = r - x. Now, the system will be in equilibrium if electrostatic force on q due to charge at A = electrostatic force on q due to charge at B.
That is, `(1)/(4pivarepsilon_(0)(qQ)/x^(2))=1/4(4pivarepsilon_(0)).(qQ)/((r-x)^(2)`
`:." "x^(2)=(r-x)^(2) implies x=(r-x)`
or`" "2x=r" "implies" "x" "=r/2`
This means that q must be placed at the centre of teo chargrs `+Q` each.
Now, the three change will be in equilibrium if the net force on each charge is zero.
i.e., `1/(4pivarepsilon_(0))(qQ)/((r//2)^(2))+1/(4pivarepsilon_(0)).("QQ")/r^(2)=0 implies" "q=-Q/4.`
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