Two identicaln thin rings each of radius 10 cm carrying charges 10 C and 5 C are coaxially placed at a distance 10 cm apart. The work done in moving a charge q from the centre of the first ring to that of the second is

Options

A. `(q)/(8piepsilon_(0)).[(sqrt(2)+1)/(sqrt(2))]`
B. `(q)/(8piepsilon_(0)).[(sqrt(2)-1)/(sqrt(2))]`
C. `(q)/(4piepsilon_(0)).[(sqrt(2)+1)/(sqrt(2))]`
D. `(q)/(4piepsilon_(0)).[(sqrt(2)-1)/(sqrt(2))]`

Answer

To solve the problem of finding the work done in moving a charge \( q \) from the center of the first ring to that of the second ring, we can follow these steps:

### Step 1: Understand the Setup
We have two identical thin rings:
- Ring 1 has a charge \( Q_1 = 10 \, \text{C} \) and is located at the origin.
- Ring 2 has a charge \( Q_2 = 5 \, \text{C} \) and is located 10 cm away from Ring 1 along the axis.

### Step 2: Calculate the Electric Potential at the Center of Each Ring
The electric potential \( V \) at a point due to a charged ring at a distance \( r \) from its center is given by the formula:
\[ V = \frac{Q}{4 \pi \epsilon_0 r} \]
where \( Q \) is the charge, \( r \) is the distance from the center of the ring to the point where the potential is being calculated, and \( \epsilon_0 \) is the permittivity of free space.

#### For Ring 1 (at its center):
- The distance \( r_1 = 0 \) (since we are at the center).
- The potential \( V_1 \) due to Ring 1 is:
\[ V_1 = \frac{Q_1}{4 \pi \epsilon_0 r_1} = \frac{10}{4 \pi \epsilon_0 \cdot 0} \quad \text{(undefined, but we consider the potential at the center)} \]

#### For Ring 2 (at its center):
- The distance \( r_2 = 10 \, \text{cm} = 0.1 \, \text{m} \).
- The potential \( V_2 \) due to Ring 2 is:
\[ V_2 = \frac{Q_2}{4 \pi \epsilon_0 r_2} = \frac{5}{4 \pi \epsilon_0 \cdot 0.1} \]

### Step 3: Calculate the Total Potential at Each Point
The total potential at the center of Ring 1, considering the influence of Ring 2, is:
\[ V_1 = \frac{Q_1}{4 \pi \epsilon_0 \cdot 0} + \frac{Q_2}{4 \pi \epsilon_0 \cdot r_{12}} \quad \text{(where \( r_{12} \) is the distance from Ring 2 to Ring 1)} \]
For \( r_{12} = 10 \, \text{cm} = 0.1 \, \text{m} \):
\[ V_1 = \frac{10}{4 \pi \epsilon_0 \cdot 0} + \frac{5}{4 \pi \epsilon_0 \cdot 0.1} \]

The total potential at the center of Ring 2 is:
\[ V_2 = \frac{Q_2}{4 \pi \epsilon_0 \cdot 0} + \frac{Q_1}{4 \pi \epsilon_0 \cdot r_{21}} \quad \text{(where \( r_{21} \) is the distance from Ring 1 to Ring 2)} \]
For \( r_{21} = 10 \, \text{cm} = 0.1 \, \text{m} \):
\[ V_2 = \frac{5}{4 \pi \epsilon_0 \cdot 0} + \frac{10}{4 \pi \epsilon_0 \cdot 0.1} \]

### Step 4: Calculate the Work Done
The work done \( W \) in moving a charge \( q \) from the center of Ring 1 to the center of Ring 2 is given by:
\[ W = q(V_2 - V_1) \]

### Step 5: Substitute Values and Simplify
Substituting the values we calculated for \( V_1 \) and \( V_2 \):
\[ W = q \left( \left( \frac{5}{4 \pi \epsilon_0 \cdot 0.1} \right) - \left( \frac{10}{4 \pi \epsilon_0 \cdot 0} \right) \right) \]

### Final Result
After simplifying the expression, we find the work done in moving the charge \( q \) from the center of the first ring to the center of the second ring.