Two metal spheres of radii `a` and `b` are connected by a thin wire. Their separation is very large compared to their dimensions. The capacitance of this system is

Options

A. `4piepsilon_0(ab)`
B. `2piepsilon_0(a+b)`
C. `4piepsilon_0(a+b)`
D. `4piepsilon_0((a^2+b^2)/2)`

Explanation

To find the capacitance of the system comprising two metal spheres of radii \( a \) and \( b \) connected by a thin wire, we can follow these steps:

### Step 1: Understand the Configuration
- We have two metal spheres with radii \( a \) and \( b \).
- They are connected by a thin wire, which means they will have the same electric potential when charged.

### Step 2: Write the Expression for Capacitance
- The capacitance \( C \) of a single sphere is given by the formula:
\[ C = 4 \pi \epsilon_0 r \]
where \( r \) is the radius of the sphere and \( \epsilon_0 \) is the permittivity of free space.

### Step 3: Calculate the Capacitance of Each Sphere
- For sphere 1 with radius \( a \):
\[ C_1 = 4 \pi \epsilon_0 a \]
- For sphere 2 with radius \( b \):
\[ C_2 = 4 \pi \epsilon_0 b \]

### Step 4: Determine the Equivalent Capacitance
- Since the two spheres are connected in parallel (they are at the same potential), the total or equivalent capacitance \( C_{eq} \) is the sum of the individual capacitances:
\[ C_{eq} = C_1 + C_2 \]
Substituting the values of \( C_1 \) and \( C_2 \):
\[ C_{eq} = 4 \pi \epsilon_0 a + 4 \pi \epsilon_0 b \]

### Step 5: Factor Out Common Terms
- We can factor out \( 4 \pi \epsilon_0 \):
\[ C_{eq} = 4 \pi \epsilon_0 (a + b) \]

### Final Result
- The capacitance of the system of two metal spheres connected by a wire is:
\[ C_{eq} = 4 \pi \epsilon_0 (a + b) \]