Question

Two conducting spheres $A$ and $B$ have large separation and connected by thin wire with a open switch. The radii of these spheres are $R$ and $2R$ respectively and initially sphere $B$ is uncharged while $A$ have charge $Q$. A student close the switch for a long time and then open and again sphere $A$ is recharge such that charge on it became same as initial. The process is repeated for many times.

Charge on sphere $B$ after three such contact with sphere $A$ is $\frac{38Q}{\alpha}$. Find $\alpha$.

Answer 1

27

Answer 2

Here's a step-by-step solution:

1. Understanding Charge Distribution:

   When two conducting spheres are connected, charge flows until their electric potentials become equal. For spheres with radii $R_A$ and $R_B$ and charges $Q_A'$ and $Q_B'$, their potentials are given by $V_A = \frac{k Q_A'}{R_A}$ and $V_B = \frac{k Q_B'}{R_B}$. Equating the potentials:

   $\frac{k Q_A'}{R_A} = \frac{k Q_B'}{R_B}$

   Given $R_A = R$ and $R_B = 2R$:

   $\frac{Q_A'}{R} = \frac{Q_B'}{2R} \implies Q_B' = 2 Q_A'$

   Also, the total charge before connection ($Q_{total}$) is conserved: $Q_A' + Q_B' = Q_{total}$. Substituting $Q_B' = 2 Q_A'$ into the conservation equation:

   $Q_A' + 2 Q_A' = Q_{total} \implies 3 Q_A' = Q_{total} \implies Q_A' = \frac{Q_{total}}{3}$

   And $Q_B' = 2 \left(\frac{Q_{total}}{3}\right) = \frac{2 Q_{total}}{3}$.

2. Tracing the Process for Each Cycle:

   Let $Q_{A,n}$ and $Q_{B,n}$ be the charges on spheres A and B just before the $n$-th contact. Let $Q'_{A,n}$ and $Q'_{B,n}$ be the charges on spheres A and B just after the $n$-th contact. After each contact, sphere A is recharged to $Q$, while sphere B retains its charge.

   Cycle 1:

   • Initial state: Sphere A has charge $Q_{A,1} = Q$. Sphere B is uncharged, so $Q_{B,1} = 0$.

   • Total charge during contact: $Q_{total,1} = Q_{A,1} + Q_{B,1} = Q + 0 = Q$.

   • Charge after contact:

     $Q'_{A,1} = \frac{Q_{total,1}}{3} = \frac{Q}{3}$

     $Q'_{B,1} = \frac{2 Q_{total,1}}{3} = \frac{2Q}{3}$

   • After opening switch and recharging A: Sphere A is recharged to $Q$. Sphere B has $Q_{B,2} = Q'_{B,1} = \frac{2Q}{3}$.

   Cycle 2:

   • Initial state: Sphere A has charge $Q_{A,2} = Q$. Sphere B has charge $Q_{B,2} = \frac{2Q}{3}$.

   • Total charge during contact: $Q_{total,2} = Q_{A,2} + Q_{B,2} = Q + \frac{2Q}{3} = \frac{5Q}{3}$.

   • Charge after contact:

     $Q'_{A,2} = \frac{Q_{total,2}}{3} = \frac{1}{3} \left(\frac{5Q}{3}\right) = \frac{5Q}{9}$

     $Q'_{B,2} = \frac{2 Q_{total,2}}{3} = \frac{2}{3} \left(\frac{5Q}{3}\right) = \frac{10Q}{9}$

   • After opening switch and recharging A: Sphere A is recharged to $Q$. Sphere B has $Q_{B,3} = Q'_{B,2} = \frac{10Q}{9}$.

   Cycle 3:

   • Initial state: Sphere A has charge $Q_{A,3} = Q$. Sphere B has charge $Q_{B,3} = \frac{10Q}{9}$.

   • Total charge during contact: $Q_{total,3} = Q_{A,3} + Q_{B,3} = Q + \frac{10Q}{9} = \frac{19Q}{9}$.

   • Charge after contact:

     $Q'_{A,3} = \frac{Q_{total,3}}{3} = \frac{1}{3} \left(\frac{19Q}{9}\right) = \frac{19Q}{27}$

     $Q'_{B,3} = \frac{2 Q_{total,3}}{3} = \frac{2}{3} \left(\frac{19Q}{9}\right) = \frac{38Q}{27}$

3. Finding $\alpha$:

   The charge on sphere B after three such contacts is $Q'_{B,3} = \frac{38Q}{27}$. The problem states this charge is $\frac{38Q}{\alpha}$. Comparing the two expressions:

   $\frac{38Q}{27} = \frac{38Q}{\alpha}$

   Therefore, $\alpha = 27$.