consider three electric charges—a, b, and c—arranged in a straight line with equal spacing between them…

Explanation:

Step1: Apply Coulomb's law

The force between two charges $q_1$ and $q_2$ separated by a distance $r$ is given by $F = k\frac{q_1q_2}{r^2}$, where $k$ is Coulomb's constant.

Step2: Calculate force on charge A

Charge A experiences a force from B and C. The force due to B is $F_{AB}=k\frac{q\times q}{d^2}$ (attractive) and due to C is $F_{AC}=k\frac{q\times q}{(2d)^2}$ (repulsive). The net force on A, $F_A = k\frac{q^2}{d^2}-k\frac{q^2}{4d^2}=k\frac{3q^2}{4d^2}$.

Step3: Calculate force on charge B

Charge B experiences a force from A and C. The force due to A is $F_{BA}=k\frac{q\times q}{d^2}$ (attractive) and due to C is $F_{BC}=k\frac{q\times q}{d^2}$ (attractive). The net force on B, $F_B=k\frac{q^2}{d^2}+k\frac{q^2}{d^2}=k\frac{2q^2}{d^2}$.

Step4: Calculate force on charge C

Charge C experiences a force from A and B. The force due to B is $F_{CB}=k\frac{q\times q}{d^2}$ (attractive) and due to A is $F_{CA}=k\frac{q\times q}{(2d)^2}$ (attractive). The net force on C, $F_C = k\frac{q^2}{d^2}+k\frac{q^2}{4d^2}=k\frac{5q^2}{4d^2}$.

Step5: Compare the magnitudes

Comparing $F_A = k\frac{3q^2}{4d^2}$, $F_B=k\frac{2q^2}{d^2}=\frac{8q^2}{4d^2}$ and $F_C = k\frac{5q^2}{4d^2}$, we have $F_A<F_C<F_B$. So the order of increasing magnitude of net - force is $A < C < B$.

Answer:

b. $A < B < C$