Question

A conducting body 1 has some initial charge $Q = 729C$, and its capacitance is C. There are two other conducting bodies, 2 and 3, having capacitance: $C_2 = 2C$ and $C_3 \rightarrow \infty$. Bodies 2 and 3 are initially uncharged. Body 2 is touched with body 1. Then, body 2 is removed from body 1 and touched with body 3, and then removed. This process is repeated for 4 times. Then, the charge on body 1 at the end must be (in C).

Answer 1

9

Answer 2

The problem describes a process of charge transfer between three conducting bodies. Let's denote the capacitances as $C_1 = C$, $C_2 = 2C$, and $C_3 \rightarrow \infty$. The initial charge on body 1 is $Q_1^{(0)} = 729 \, C$, while bodies 2 and 3 are initially uncharged. The process is repeated 4 times.

Let's analyze one cycle of the process:

A. Body 2 is touched with body 1.

When two conducting bodies are touched, charge redistributes until they reach the same potential. The total charge is conserved.
Let $Q_1^{(n-1)}$ be the charge on body 1 at the beginning of the $n$-th cycle, and body 2 is uncharged ($Q_2 = 0$).
After touching, the new charges $Q_1'$ and $Q_2'$ will be:

$Q_1' = \frac{C_1}{C_1 + C_2} (Q_1^{(n-1)} + Q_2)$

$Q_2' = \frac{C_2}{C_1 + C_2} (Q_1^{(n-1)} + Q_2)$

Substituting the given values $C_1 = C$, $C_2 = 2C$, and $Q_2 = 0$:

$Q_1' = \frac{C}{C + 2C} Q_1^{(n-1)} = \frac{C}{3C} Q_1^{(n-1)} = \frac{1}{3} Q_1^{(n-1)}$

$Q_2' = \frac{2C}{C + 2C} Q_1^{(n-1)} = \frac{2C}{3C} Q_1^{(n-1)}$

So, after this step, body 1 has charge $\frac{1}{3} Q_1^{(n-1)}$ and body 2 has charge $\frac{2}{3} Q_1^{(n-1)}$.

B. Body 2 is removed from body 1 and touched with body 3.

Body 2 now carries charge $Q_2' = \frac{2}{3} Q_1^{(n-1)}$. Body 3 has infinite capacitance ($C_3 \rightarrow \infty$). When a charged body is touched to a body with infinite capacitance (like grounding), all its charge flows to the body with infinite capacitance, as its potential effectively remains zero.
Therefore, after touching body 3, the charge on body 2 becomes $0$. The charge $Q_2'$ is transferred to body 3.
The charge on body 1 is unaffected by this step, as body 2 has already been removed from it.

So, at the end of one complete cycle, the charge on body 1, $Q_1^{(n)}$, is:

$Q_1^{(n)} = Q_1' = \frac{1}{3} Q_1^{(n-1)}$

This means that in each cycle, the charge on body 1 becomes one-third of its charge at the beginning of that cycle.

The process is repeated for 4 times. We can find the charge on body 1 after 4 cycles:

$Q_1^{(1)} = \frac{1}{3} Q_1^{(0)}$

$Q_1^{(2)} = \frac{1}{3} Q_1^{(1)} = \left(\frac{1}{3}\right)^2 Q_1^{(0)}$

$Q_1^{(3)} = \frac{1}{3} Q_1^{(2)} = \left(\frac{1}{3}\right)^3 Q_1^{(0)}$

$Q_1^{(4)} = \frac{1}{3} Q_1^{(3)} = \left(\frac{1}{3}\right)^4 Q_1^{(0)}$

Given initial charge $Q_1^{(0)} = 729 \, C$.

$Q_1^{(4)} = \left(\frac{1}{3}\right)^4 \times 729$

We know that $3^4 = 81$ and $729 = 3^6$.

$Q_1^{(4)} = \frac{1}{81} \times 729 = \frac{3^6}{3^4} = 3^{(6-4)} = 3^2 = 9 \, C$.

The charge on body 1 at the end of 4 repetitions is $9 \, C$.

Explanation of the solution:

1. Charge Distribution (Body 1 & 2): When body 1 (charge $Q_1$, capacitance $C$) and body 2 (uncharged, capacitance $2C$) are touched, the total charge $Q_1$ redistributes. The charge on body 1 becomes $Q_1' = \frac{C}{C+2C} Q_1 = \frac{1}{3} Q_1$. The charge on body 2 becomes $Q_2' = \frac{2C}{C+2C} Q_1 = \frac{2}{3} Q_1$.
2. Charge Transfer (Body 2 & 3): Body 2, now charged with $Q_2'$, is touched to body 3 (infinite capacitance). All its charge $Q_2'$ transfers to body 3, leaving body 2 uncharged ($0$). Body 1's charge remains $Q_1'$.
3. Recursive Relation: After one cycle, the charge on body 1 is $Q_1' = \frac{1}{3} Q_1$. This means the charge on body 1 is reduced by a factor of 3 in each cycle.
4. Calculation: After 4 cycles, the charge on body 1 will be $Q_1^{(4)} = \left(\frac{1}{3}\right)^4 Q_1^{(0)}$. Given $Q_1^{(0)} = 729 \, C$, $Q_1^{(4)} = \left(\frac{1}{3}\right)^4 \times 729 = \frac{1}{81} \times 729 = 9 \, C$.