Answer:
[tex]q_1=+0.375\ {10}^{-9}[/tex]
Explanation:
Electrostatic Forces
The force exerted between two point charges [tex]q_1[/tex] and [tex]q_2[/tex] separated a distance d is given by Coulomb's formula
[tex]\displaystyle F=\frac{k\ q_1\ q_2}{d^2}[/tex]
The forces are attractive if the charges have different signs and repulsive if they have equal signs.
The problem described in the question locates three point charges in a straight line. The charges have the values shown below
[tex]\displaystyle q_3=+5\ 10^{-9}\ c[/tex]
[tex]\displaystyle q_2=-3\ 10^{-9}\ c[/tex]
The distance between [tex]q_3[/tex] and [tex]q_2[/tex] is
[tex]\displaystyle d_2=4cm=0.04\ m[/tex]
The distance between [tex]q_3[/tex] and [tex]q_1[/tex] is
[tex]\displaystyle d_1=2cm=0.02\ m[/tex]
We must find the value of [tex]q_1[/tex] such that
[tex]\displaystyle |F_3|=0[/tex]
Applying Coulomb's formula for [tex]q_1[/tex] is
[tex]\displaystyle F_{13}=\frac{k\ q_1\ q_3}{d_1^2}[/tex]
Now for [tex]q_2[/tex]
[tex]\displaystyle F_{23}=\frac{k\ q_2\ q_3}{d_2^2}[/tex]
If the total force on [tex]q_3[/tex] is zero, both forces must be equal. Note that being q2 negative, the force on q3 is to the right. The force exerted by q1 must go to the left, thus q1 must be positive. Equating the forces we have:
[tex]\displaystyle F_{13}=F_{23}[/tex]
[tex]\displaystyle \frac{k\ q_1\ q_3}{d_1^2}=\frac{k\ q_2\ q_3}{d_2^2}[/tex]
Simplfying and solving for [tex]q_1[/tex]
[tex]\displaystyle q_1=\frac{q_2\ d_1^2}{d_2^2}[/tex]
[tex]\displaystyle q_1=\frac{3.10^{-9}\ 0.02^2}{0.04^2}[/tex]
[tex]\boxed{\displaystyle q_1=+0.375\ {10}^{-9}}[/tex]
