Answer :
Answer:
[tex]+1.11\mu C[/tex]
Explanation:
A charge located at a point will experience a zero electrostatic force if the resultant electric field on it due to any other charge(s) is zero.
[tex]Q_1[/tex] is located at the origin. The net force on it will only be zero if the resultant electric field intensity due to [tex]Q_2[/tex] and [tex]Q_3[/tex] at the origin is equal to zero. Therefore we can perform this solution without necessarily needing the value of [tex]Q_1[/tex].
Let the electric field intensity due to [tex]Q_2[/tex] be +[tex]E_2[/tex] and that due to [tex]Q_3[/tex] be -[tex]E_3[/tex] since the charge is negative. Hence at the origin;
[tex]+E_2-E_3=0..................(1)[/tex]
From equation (1) above, we obtain the following;
[tex]E_2=E_3.................(2)[/tex]
From Coulomb's law the following relationship holds;
[tex]+E_2=\frac{kQ_2}{r_2^2}\\[/tex]
[tex]-E_3=\frac{kQ_3}{r_3^2}[/tex]
where [tex]r_2[/tex] is the distance of [tex]Q_2[/tex] from the origin, [tex]r_3[/tex] is the distance of [tex]Q_3[/tex] from the origin and k is the electrostatic constant.
It therefore means that from equation (2) we can write the following;
[tex]\frac{kQ_2}{r_2^2}=\frac{kQ_3}{r_3^2}.................(3)[/tex]
k can cancel out from both side of equation (3), so that we finally obtain the following;
[tex]\frac{Q_2}{r_2^2}=\frac{Q_3}{r_3^2}................(4)[/tex]
Given;
[tex]Q_2=?\\r_2=2.5cm=0.025m\\Q_3=-2.18\mu C=-2.18* 10^{-6}C\\r_3=3.5cm=0.035m[/tex]
Substituting these values into equation (4); we obtain the following;
[tex]\frac{Q_2}{0.025^2}=\frac{2.18*10^{-6}}{0.035^2}\\\\hence;\\\\Q_2=\frac{0.025^2*2.18*10^{-6}}{0.035^2}\\[/tex]
[tex]Q_2=\frac{0.00136*10^{-6}}{0.00123}=1.11*10^{-6}C\\\\Q_3=+1.11\mu C[/tex]