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

Explanation:

Step1: Apply Coulomb's law

Coulomb's law is $F = k\frac{q_1q_2}{r^2}$, where $k$ is Coulomb's constant, $q_1$ and $q_2$ are charges, and $r$ is the distance between them.

Step2: Calculate force on charge A

The force on charge A due to B is $F_{AB}=k\frac{q\times q}{d^2}$ (attractive, to the right) and due to C is $F_{AC}=k\frac{q\times q}{(2d)^2}$ (attractive, to the right). Net force on A, $F_A=kq^2(\frac{1}{d^2}+\frac{1}{4d^2})=k\frac{5q^2}{4d^2}$.

Step3: Calculate force on charge B

The force on charge B due to A is $F_{BA}=k\frac{q\times q}{d^2}$ (attractive, to the left) and due to C is $F_{BC}=k\frac{q\times q}{d^2}$ (repulsive, to the left). Net force on B, $F_B = k\frac{2q^2}{d^2}$.

Step4: Calculate force on charge C

The force on charge C due to A is $F_{CA}=k\frac{q\times q}{(2d)^2}$ (attractive, to the left) and due to B is $F_{CB}=k\frac{q\times q}{d^2}$ (repulsive, to the left). Net force on C, $F_C=kq^2(\frac{1}{d^2}+\frac{1}{4d^2})=k\frac{5q^2}{4d^2}$.

Step5: Compare forces

Comparing magnitudes: $F_A = F_C=\frac{5kq^2}{4d^2}$ and $F_B=\frac{2kq^2}{d^2}=\frac{8kq^2}{4d^2}$. So $F_A = F_C<F_B$, or $A = C<B$.

Answer:

D. B>A=C