Question

consider three electric charges—a, b, and c—arranged in a straight line with equal spacing between them shown in the figure below. rank the charges in order of increasing magnitude of the net force they experience. indicate ties where appropriate. a. b = c = a b. a < b < c c. c < a < b d. b < a < c e. b = a < a

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

Answer:

D. B>A=C