Respuesta :
Answer:
[tex]\rm |q_1|=8.0\times 10^{-7}\ C,\ \ \ |q_2| = 4.6\times 10^{-6}\ C.[/tex]
Explanation:
According to the Coulomb's law, the magnitude of the electrostatic force between two static point charges [tex]\rm q_1[/tex] and [tex]\rm q_1[/tex], separated by a distance [tex]\rm r[/tex], is given by
[tex]\rm F = \dfrac{kq_1q_2}{r^2}.[/tex]
where k is the Coulomb's constant.
Initially,
[tex]\rm r = 0.160\ m\\F_i = -1.30\ N.\\\\and \ \ |q_2|>|q_1|.[/tex]
The negative sign is taken with force F because the force is attractive.
Therefore, the initial electrostatic force between the charges is given by
[tex]\rm F_i = \dfrac{kq_1q_2}{r^2}.\\-1.30=\dfrac{kq_1q_2}{0.160^2}\\\rm\Rightarrow q_2 = \dfrac{-1.30\times 0.160^2}{q_1k}\ \ \ ..............\ (1).[/tex]
Now, the objects are then brought into contact, so the net charge is shared equally, and then they are returned to their initial positions.
The force is now repulsive, therefore, [tex]\rm F_f = +1.30\ N.[/tex]
The new charges on the two objects are
[tex]\rm q_1'=q_2' = \dfrac{q_1+q_2}{2}.[/tex]
The new force is given by
[tex]\rm F_f = \dfrac{kq_1'q_2'}{r^2}\\+1.30=\dfrac{k\left (\dfrac{q_1+q_2}{2}\right )\left (\dfrac{q_1+q_2}{2}\right )}{0.160^2}\\\Rightarrow \left (\dfrac{q_1+q_2}{2}\right )^2=\dfrac{+1.30\times 0.160^2}{k}\\(q_1+q_2)^2=\dfrac{4\times 1.30\times 0.160^2}{k}\\q_1^2+q_2^2+2q_1q_2=\dfrac{4\times 1.30\times 0.160^2}{k}\\\\[/tex]
Using (1),
[tex]\rm q_1^2+\left ( \dfrac{-1.30\times 0.160^2}{q_1k}\right )^2+2\left (\dfrac{-1.30\times 0.160^2}{k} \right )=\dfrac{4\times 1.30\times 0.160^2}{k}\\q_1^2+\dfrac 1{q_1^2}\left ( \dfrac{-1.30\times 0.160^2}{k}\right )^2-\left (\dfrac{6\times 1.30\times 0.160^2}{k} \right )=0\\q_1^4+\left ( \dfrac{-1.30\times 0.160^2}{k}\right )^2-q_1^2\left (\dfrac{6\times 1.30\times 0.160^2}{k} \right )=0[/tex]
[tex]\rm q_1^4+\left ( \dfrac{-1.30\times 0.160^2}{k}\right )^2-q_1^2\left (\dfrac{6\times 1.30\times 0.160^2}{k} \right )=0\\q_1^4+\left ( \dfrac{-1.30\times 0.160^2}{9\times 10^9}\right )^2-q_1^2\left (\dfrac{6\times 1.30\times 0.160^2}{9\times 10^9} \right )=0\\q_1^4-q_1^2\left (\dfrac{6\times 1.30\times 0.160^2}{9\times 10^9} \right )+\left ( \dfrac{-1.30\times 0.160^2}{9\times 10^9}\right )^2=0[/tex]
[tex]\rm q_1^4-q_1^2\left (2.22\times 10^{-11} \right )+\left ( 1.37\times 10^{-23}\right ) =0\\\Rightarrow q_1^2 = \dfrac{-(-2.22\times 10^{-11})\pm \sqrt{(-2.22\times 10^{-11})^2-4\cdot (1)\cdot (1.37\times 10^{-23})}}{2}\\=1.11\times 10^{-11}\pm 1.046\times 10^{-11}.\\=6.4\times 10^{-13}\ \ \ or\ \ \ 2.156\times 10^{-11}\\\Rightarrow q_1 = \pm 8.00\times 10^{-7}\ C\ \ \ or\ \ \ \pm 4.64\times 10^{-6}\ C.[/tex]
Using (1),
When [tex]\rm q_1 = \pm 8.00\times 10^{-7}\ C[/tex],
[tex]\rm q_2=\dfrac{-1.30\times 0.160^2}{\pm 8.00\times 10^{-7}\times 9\times 10^9}=\mp4.6\times 10^{-6}\ C.[/tex]
When [tex]\rm q_1=\pm 4.6\times 10^{-6}\ C[/tex],
[tex]\rm q_2=\dfrac{-1.30\times 0.160^2}{\pm 4.64\times 10^{-6}\times 9\times 10^9}=\mp7.97\times 10^{-7}\ C\approx 8.0\times 10^{-7}\ C.[/tex]
Since, [tex]\rm |q_2|>|q_1|[/tex]
Therefore, [tex]\rm |q_1|=8.0\times 10^{-7}\ C,\ \ \ |q_2| = 4.6\times 10^{-6}\ C.[/tex]
