Question

four pairs of conducting spheres, all with the same radius, are shown in the figure below. each sphere has a net charge placed on it initially. the spheres in each pair are then brought into contact, allowing charge to redistribute between them. rank the pairs of spheres in order of increasing magnitude of the charge transferred during contact. indicate ties where appropriate.

a. c<a<d<b
b. c<d<a<b
c. b<a<c<d
d. d<a<c<b
e. a<d<b<c

Explanation:

Step1: Recall charge - sharing formula

When two conducting spheres of the same radius are brought in contact, the final charge on each sphere is $q_{final}=\frac{q_1 + q_2}{2}$, and the charge transferred $\Delta q$ is related to the initial and final charges.

Step2: Calculate charge transfer for pair A

Initial charges are $q_1=-q$ and $q_2 = - 5q$. Final charge on each sphere is $q_{final}=\frac{-q-5q}{2}=-3q$. Charge transferred $\Delta q_A=\vert -q-(-3q)\vert = 2q$.

Step3: Calculate charge transfer for pair B

Initial charges are $q_1 = 0$ and $q_2=+8q$. Final charge on each sphere is $q_{final}=\frac{0 + 8q}{2}=4q$. Charge transferred $\Delta q_B=\vert0 - 4q\vert=4q$.

Step4: Calculate charge transfer for pair C

Initial charges are $q_1=+q$ and $q_2=-5q$. Final charge on each sphere is $q_{final}=\frac{q-5q}{2}=-2q$. Charge transferred $\Delta q_C=\vert+q-(-2q)\vert = 3q$.

Step5: Calculate charge transfer for pair D

Initial charges are $q_1=+15q$ and $q_2=+13q$. Final charge on each sphere is $q_{final}=\frac{15q + 13q}{2}=14q$. Charge transferred $\Delta q_D=\vert15q-14q\vert = q$.

Step6: Rank the charge - transfer magnitudes

Comparing $\Delta q_A = 2q$, $\Delta q_B=4q$, $\Delta q_C = 3q$, $\Delta q_D=q$, we get $\Delta q_D<\Delta q_A<\Delta q_C<\Delta q_B$.

Answer:

d. D<A<C<B