Two conducting spheres of radii r and 2r are placed at very large separation. Each sphere possesses charge Q. These spheres are connected with a conducting wire of resistance R. Then, which of the following is true?

Options

A. Initial current is `Q/(8 pi in_0 rR)`
B. Initial current is `Q/(4 pi in_0 rR)`
C. Current reduces to half the initial current after time `8pi in_0 rR ln 2`
D. Current reduces to half the initial current after time `(8pi in_0 rR ln 2)/3`

Answer

To solve the problem of two conducting spheres of radii \( r \) and \( 2r \) connected by a wire of resistance \( R \), we will follow these steps:

### Step 1: Determine the Initial Potentials of the Spheres
The potential \( V \) of a charged sphere is given by the formula:
\[ V = \frac{kQ}{r} \]
where \( k = \frac{1}{4\pi\epsilon_0} \).

For Sphere A (radius \( r \)):
\[ V_A = \frac{kQ}{r} \]

For Sphere B (radius \( 2r \)):
\[ V_B = \frac{kQ}{2r} \]

### Step 2: Calculate the Potential Difference
The potential difference \( \Delta V \) between the two spheres is:
\[ \Delta V = V_A - V_B = \frac{kQ}{r} - \frac{kQ}{2r} \]
This simplifies to:
\[ \Delta V = \frac{kQ}{r} \left(1 - \frac{1}{2}\right) = \frac{kQ}{r} \cdot \frac{1}{2} = \frac{kQ}{2r} \]

### Step 3: Calculate the Initial Current
Using Ohm's law, the current \( I \) flowing through the wire can be calculated as:
\[ I = \frac{\Delta V}{R} = \frac{\frac{kQ}{2r}}{R} = \frac{kQ}{2rR} \]

Substituting \( k = \frac{1}{4\pi\epsilon_0} \):
\[ I = \frac{Q}{8\pi\epsilon_0 r R} \]

### Step 4: Determine the Capacitance of the Spheres
The capacitance \( C \) of a conducting sphere is given by:
\[ C = 4\pi\epsilon_0 r \]
For Sphere A:
\[ C_1 = 4\pi\epsilon_0 r \]
For Sphere B:
\[ C_2 = 4\pi\epsilon_0 (2r) = 8\pi\epsilon_0 r \]

### Step 5: Calculate the Equivalent Capacitance
Since the spheres are connected in series, the equivalent capacitance \( C_{eq} \) is given by:
\[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} \]
Substituting the values:
\[ \frac{1}{C_{eq}} = \frac{1}{4\pi\epsilon_0 r} + \frac{1}{8\pi\epsilon_0 r} = \frac{2 + 1}{8\pi\epsilon_0 r} = \frac{3}{8\pi\epsilon_0 r} \]
Thus,
\[ C_{eq} = \frac{8\pi\epsilon_0 r}{3} \]

### Step 6: Determine the Time for Current to Become Half
Using the formula for current decay in a capacitor:
\[ I(t) = I_0 e^{-\frac{t}{RC}} \]
Setting \( I(t) = \frac{I_0}{2} \):
\[ \frac{I_0}{2} = I_0 e^{-\frac{t}{RC}} \]
Dividing both sides by \( I_0 \):
\[ \frac{1}{2} = e^{-\frac{t}{RC}} \]
Taking the natural logarithm:
\[ -\ln(2) = -\frac{t}{RC} \implies t = RC \ln(2) \]
Substituting \( C_{eq} \):
\[ t = R \cdot \frac{8\pi\epsilon_0 r}{3} \ln(2) \]

### Conclusion
The initial current \( I \) is given by:
\[ I = \frac{Q}{8\pi\epsilon_0 r R} \]
And the time \( t \) at which the current becomes half is:
\[ t = R \cdot \frac{8\pi\epsilon_0 r}{3} \ln(2) \]