Answer:
a. [tex]q_C[/tex] experiences the greatest net force and [tex]q_B[/tex] experiences the smallest net force
b. Ratio of the greatest to the smallest net force= 9
Explanation:
Electrostatic Forces
Two point-charges q1 and q2, separated a distance d, exert on each other an electrostatic force of magnitude
[tex]\displaystyle F=K\frac{q_1q_2}{d^2}[/tex]
If the charges have the same sign, they repel each other, for different signed charges, they attract. That gives us the direction of each force in the space.
Let's assume all the charges of the problem have a magnitude q, and between two consecutive charges, the distance is d. The proposed layout is shown it the image.
a.
The net force on qA is the sum of those exerted by qB, qC, and qD. But note qB and qC repel qA and qD attracts it, so the total force on qA is
[tex]F_{TA}=-F_B-F_C+F_D[/tex]
Computing the individual forces we have
[tex]\displaystyle F_B=\frac{K\ q_A\ q_B}{d^2}=K\ \frac{q^2}{d^2}[/tex]
[tex]\displaystyle F_C=\frac{K\ q_A\ q_C}{(2d)^2}=\frac{1}{4}\ \ \frac{K\ q^2}{d^2}[/tex]
[tex]\displaystyle F_D=\frac{K\ q_A\ q_D}{(3d)^2}=\frac{1}{9}\ \ \frac{K\ q^2}{d^2}[/tex]
The total force on qA is:
[tex]\displaystyle F_{TA}=\frac{K\ q^2}{d^2}(-1-\frac{1}{4}+\frac{1}{9})[/tex]
[tex]\displaystyle F_{TA}=-\frac{41}{36}\ \frac{K\ q^2}{d^2}[/tex]
[tex]\displaystyle |F_{TA}|=\frac{41}{36}\ \frac{K\ q^2}{d^2}[/tex]
Charge qA repels qB to the right, qC repels qB to the left, and qD attracts qB to the right, thus
[tex]\displaystyle F_{TB}=F_A-F_C+F_D[/tex]
[tex]\displaystyle F_{TB}=\frac{K\ q^2}{d^2}-\frac{K\ q^2}{d^2}+\frac{K\ q^2}{(2d)^2}[/tex]
[tex]\displaystyle F_{TB}=\frac{1}{4}\ \frac{K\ q^2}{d^2}[/tex]
[tex]\displaystyle |F_{TB}|=\frac{1}{4}\ \frac{K\ q^2}{d^2}[/tex]
Charges qA and qb repel qC to the right, and qD attracts qC to the right, thus
[tex]\displaystyle F_{TC}=F_A+F_B+F_D[/tex]
[tex]\displaystyle F_{TC}=\frac{K\ q^2}{(2d)^2}+\frac{K\ q^2}{d^2}+\frac{K\ q^2}{d^2}[/tex]
[tex]\displaystyle F_{TC}=\frac{9}{4}\ \frac{K\ q^2}{d^2}[/tex]
[tex]\displaystyle |F_{TC}|=\frac{9}{4}\ \frac{K\ q^2}{d^2}[/tex]
Charge qA and qB attract qD to the left, and qC atracts qD to the left, thus
[tex]\displaystyle F_{TD}=-F_A-F_B-F_C[/tex]
[tex]\displaystyle F_{TD}=-\frac{K\ q2}{(3d)^2}-\frac{K\ q2}{(2d)^2}-\frac{K\ q2}{d^2}[/tex]
[tex]\displaystyle F_{TD}=-\frac{49}{36}\ \frac{K\ q^2}{d^2}[/tex]
[tex]\displaystyle |F_{TD}|=\frac{49}{36}\ \frac{K\ q^2}{d^2}[/tex]
Comparing the relative values of all the forces
[tex]\displaystyle |F_{TC}|>|F_{TD}|>|F_{TA}|>|F_{TB}|[/tex]
This means that qc experiences the greatest net force and qB experiences the smallest net force
b.
The ratio of the greatest to the smallest forces is
[tex]\displaystyle \frac{|F_{TC}|}{|F_{TB}|}=\frac{\frac{9}{4}}{\frac{1}{4}}=9[/tex]
