Question

An electric point charge of Q_1 = 9.63 nC is placed at the origin of the real axis. Another point charge of Q_2 = 1.83 nC is placed at a position of p = 2.25 m on the_real axis. At which position can a third point charge of q = -7.28 nC be placed so that the net electrostatic force on it is zero? Let the sign of Q_2 be changed from positive to negative. At which position can the point charge q be placed now so that the net electrostatic force on it is zero?

Answer 1

2.25 m Q1 Q2 Q3 x m

How, Q1 and Q2 are positive, the charge Q3 is negative then this must be between these two charges for that the net force to be zero.

$F_{13}=k\frac{q_{1}q_{3}}{x^{2}}$

$F_{23}=k\frac{q_{2}q_{3}}{(2.25-x)^{2}}$

$k\frac{q_{1}q_{3}}{x^{2}}=k\frac{q_{2}q_{3}}{(2.25-x)^{2}}\rightarrow \frac{q_{1}}{x^{2}}=\frac{q_{2}}{(5.06-4.5x+x^{2})}$

$q_{1}(5.06-4.5x+x^{2})=q_{2}x^{2}\rightarrow$

$(9.63\times 10^{-9}C)(5.06-4.5x+x^{2})=(1.83\times 10^{-9}C)x^{2}$

$(9.63)(5.06-4.5x+x^{2})=(1.83)x^{2}\rightarrow 48.72-43.33x+9.63x^{2}=1.83x^{2}$

$48.72-43.33x+7.8x^{2}=0$ Now solve the quadratic equation $\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$

$\frac{-(-43.33)\pm \sqrt{(-43.33)^{2}-4(7.8)(48.72)}}{2(7.8)}$

$\frac{43.33\pm \sqrt{357.42}}{15.6}\rightarrow$

$\frac{43.33+\sqrt{357.42}}{15.6}\rightarrow x=3.98$

$\frac{43.33-\sqrt{357.42}}{15.6}\rightarrow x=1.56$

The correct option is 1.56 m then the charge Q3 must be at 1.56 m of Q1 and 0.68 m from Q2

How, Q2 and Q3 are negative, the charge Q3 must be outside these two charges, but more close of the Q2 for that the net force to be zero.

$F_{13}=k\frac{q_{1}q_{3}}{(2.25+x)^{2}}$

$F_{23}=k\frac{q_{2}q_{3}}{x^{2}}$

$k\frac{q_{2}q_{3}}{x^{2}}=k\frac{q_{1}q_{3}}{(2.25+x)^{2}}\rightarrow \frac{q_{2}}{x^{2}}=\frac{q_{1}}{(5.06+4.5x+x^{2})}$

$q_{2}(5.06+4.5x+x^{2})=q_{1}x^{2}\rightarrow$

$(1.83\times 10^{-9}C)(5.06+4.5x+x^{2})=(9.63\times 10^{-9}C)x^{2}$

$(1.83)(5.06+4.5x+x^{2})=(9.63)x^{2}\rightarrow 9.25+8.23x+1.83x^{2}=9.63x^{2}$

$9.25+8.23x-7.8x^{2}=0$ Now solve the quadratic equation $\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$

$\frac{-(8.23)\pm \sqrt{(8.23)^{2}-4(-7.8)(9.25)}}{2(-7.8)}$

$\frac{-8.23\pm \sqrt{356.33}}{-15.6}\rightarrow$

$\frac{-8.23+\sqrt{356.33}}{-15.6}\rightarrow x=-0.68m$

$\frac{-8.23-\sqrt{356.33}}{-15.6}\rightarrow x=1.73m$

The correct option is 1.73 m then the charge Q3 must be at 1.73 m of Q2 and 3.98 m from Q1

If you have any question please let me know in the comments