Respuesta :
Answer:
Largest
[q1=+1nC, q2=-1nC, q3=-1nC] & [q1=-1nC, q2=+1nC, q3=+1nC]
[q1=+1nC, q2=-1nC, q3=+1nC]
[q1=+1nC, q2=+1nC, q3=-1nC]
Smallest
[q1=+1nC, q2=+1nC, q3=+1nC] & [q1=-1nC, q2=-1nC, q3=-1nC]
Explanation:
The force between two electrical charges is directly proportional to the product of charges and inversely proportional to the square of distance between them. The ranking of the combinations of charges on the electric force acting on q1 is F2=F3 > F5 > F4> F6=F1.
Coulomb's law:
It states that the force between two electrical charges is directly proportional to the product of charges and inversely proportional to the square of distance between them.
[tex]\bold { F = k \times \frac{q_1q__2}{r^2} }[/tex]
Where,
q1 and q2 - charges
r - distance
k - Coulomb's constant = [tex]\bold{8.99\times10^9Nm^2C^-^2}[/tex]
Given here,
3 charges q1, q2, and q3
Case 1
q1 = q2= q3= +1nC
All charges have same sign, hence the force between them is repulsive.
Force due to q2 on q1,
[tex]\bold{F_2_1 = k \frac{q1q2}{r2} = -k \frac{1}{d^2} nN}[/tex]
Force due to q3 on q1,
[tex]\bold{F_3_1 = k \frac{q1q3}{r2} = -k \frac{1}{4d^2} nN}[/tex]
The net force on q1
[tex]\bold{F1=F_2_1+F_3_1}\\\\\bold{F1=-k\frac{1}{d^2} +(-k\frac{1}{4d^2} )}\\\\\bold{F1=-\frac{5}{4} (\frac{k}{d^2} ) nN}}[/tex]
Using the formula above, the force can be calculated for other cases.
Case-2
[tex]\bold{F2=\frac{5}{4} (\frac{k}{d^2} ) nN}}[/tex]
Case-3
[tex]\bold{F3=\frac{5}{4} (\frac{k}{d^2} ) nN}}[/tex]
Case-4
[tex]\bold{F4=-\frac{3}{4} (\frac{k}{d^2} ) nN}}[/tex]
Case-5
[tex]\bold{F5=\frac{3}{4} (\frac{k}{d^2} ) nN}}[/tex]
Case-6
[tex]\bold{F6=-\frac{5}{4} (\frac{k}{d^2} ) nN}}[/tex]
Therefore, the ranking of the combinations of charges on the electric force acting on q1 is F2=F3 > F5 > F4> F6=F1.
To know more about Coulomb's law, refer to the link:
https://brainly.com/question/1698562
