Question

Can anyone help me with the third part?? Thanks so much!

Three charges Q1, Q2 , and Q3 are located in a straight line. The position of Q2 is 0.235 m to the right Q1.Q3 is located 0.169 m to the right of Q2. The force on Q2 due to its interaction with Q-3 is directed to the..... True: Right if the two charges have opposite signs. False: Left if the two charges have opposite signs.

Answer 1

Q1 -Q2 X12 x23

The figure show the arrangement of the charges. As you can see we want to find x. The net force on Q3 is

$F=k_e\frac{Q_1Q_3}{x^2}+k_e\frac{-Q_2Q_3}{x_{23}\,^2}$

Like we do not know the distance from Q2 to Q3 we go tu use the relation

$x=x_{12}+x_{23}$

$x_{23}=x-x_{12}$

Then

$F=k_e\frac{Q_1Q_3}{x^2}+k_e\frac{-Q_2Q_3}{(x-x_{12})\,^2}$

$F=\left(\frac{Q_1}{x^2}-\frac{Q_2}{(x-x_{12})\,^2} \right )k_eQ_3$

Like the net force must be zero

$0=\left(\frac{Q_1}{x^2}-\frac{Q_2}{(x-x_{12})\,^2} \right )k_eQ_3$

$0=\frac{Q_1}{x^2}-\frac{Q_2}{(x-x_{12})\,^2}$

$0=\frac{Q_1(x-x_{12})\,^2-Q_2(x^2)}{x^2(x-x_{12})\,^2}$

$0={Q_1(x-x_{12})\,^2-Q_2(x^2)}$

$0=Q_1(x^2-2x_{12}x+x_{12}\,^2)-Q_2x^2$

$0=Q_1x^2-2Q_1x_{12}x+Q_1x_{12}\,^2-Q_2x^2$

$0=(Q_1-Q_2)x^2-2Q_1x_{12}x+Q_1x_{12}\,^2$

The solution of this equation is given by

$x=\frac{-(-2Q_1x_{12})\pm\sqrt{(-2Q_1x_{12})^2-4(Q_1-Q_2)(Q_1x_{12}\,^2)}}{2(Q_1-Q_2)}$

$x=\frac{2Q_1x_{12}\pm\sqrt{4Q_1\,^2x_{12}\,^2-4Q_1\,^2x_{12}\,^2+4Q_2Q_1x_{12}\,^2}}{2(Q_1-Q_2)}$

$x=\frac{2Q_1x_{12}\pm\sqrt{4Q_2Q_1x_{12}\,^2}}{2(Q_1-Q_2)}$

$x=\frac{2Q_1x_{12}\pm2\sqrt{Q_2Q_1}x_{12}}{2(Q_1-Q_2)}$

$x=\frac{Q_1\pm\sqrt{Q_2Q_1}}{Q_1-Q_2}x_{12}$

$x=\frac{(1.56\times 10^{-6}C)\pm\sqrt{(2.84\times 10^{-6}C)(1.56\times 10^{-6}C)}}{(1.56\times 10^{-6}C)-(2.84\times 10^{-6}C)}(0.235m)$

Here we have two solutions

$x_1=-0.673m$  at the left of the charge 1

$x_2=0.100m$ at the right of the charge 1