Among two discs A and B, first have radius 10 cm and charge `10^(-6) mu`C and second have radius 30 cm and charge `10^(-5)mu` C. When they are touched, charge on both `q_(A)` and` q_(B)` respectively will, be

Options

A. `q_(A) = 2.75 muC, q_(B) = 3.15 muC`
B. `q_(A) = 1.09 muC, q_(B) 1.53 muC`
C. `q_(A) = q_(B) = 5.5 muC`
D. None of these

Answer

To solve the problem, we need to find the charges on two discs A and B after they are touched. The discs have different radii and initial charges. We will follow these steps:

### Step 1: Identify the given data
- Charge on disc A, \( q_A = 10^{-6} \, \mu C = 1 \, \mu C \)
- Charge on disc B, \( q_B = 10^{-5} \, \mu C = 10 \, \mu C \)
- Radius of disc A, \( r_A = 10 \, cm = 0.1 \, m \)
- Radius of disc B, \( r_B = 30 \, cm = 0.3 \, m \)

### Step 2: Use the formula for electric potential
When the discs are touched, they will share their charges until the electric potential on both discs is equal. The electric potential \( V \) due to a charged disc is given by:

\[ V = k \frac{q}{r} \]

where \( k \) is the electrostatic constant.

### Step 3: Set the potentials equal
Since the potentials will be equal after touching:

\[ V_A = V_B \]

This gives us:

\[ k \frac{q_A'}{r_A} = k \frac{q_B'}{r_B} \]

where \( q_A' \) and \( q_B' \) are the final charges on discs A and B, respectively.

### Step 4: Cancel out the constant \( k \)
We can cancel \( k \) from both sides:

\[ \frac{q_A'}{r_A} = \frac{q_B'}{r_B} \]

### Step 5: Express \( q_A' \) in terms of \( q_B' \)
Rearranging gives us:

\[ \frac{q_A'}{q_B'} = \frac{r_A}{r_B} \]

Substituting the values of \( r_A \) and \( r_B \):

\[ \frac{q_A'}{q_B'} = \frac{0.1}{0.3} = \frac{1}{3} \]

This means:

\[ q_A' = \frac{1}{3} q_B' \]

### Step 6: Find the total charge
The total charge before they are touched is:

\[ q_{total} = q_A + q_B = 1 \, \mu C + 10 \, \mu C = 11 \, \mu C \]

### Step 7: Set up the equation for total charge
Now we can express the total charge in terms of \( q_A' \) and \( q_B' \):

\[ q_A' + q_B' = 11 \, \mu C \]

Substituting \( q_A' \):

\[ \frac{1}{3} q_B' + q_B' = 11 \, \mu C \]

### Step 8: Solve for \( q_B' \)
Combining terms gives:

\[ \frac{4}{3} q_B' = 11 \, \mu C \]

Multiplying both sides by \( \frac{3}{4} \):

\[ q_B' = \frac{3}{4} \times 11 \, \mu C = 8.25 \, \mu C \]

### Step 9: Find \( q_A' \)
Now substituting back to find \( q_A' \):

\[ q_A' = \frac{1}{3} q_B' = \frac{1}{3} \times 8.25 \, \mu C = 2.75 \, \mu C \]

### Final Charges
Thus, the final charges on the discs after they are touched are:
- Charge on disc A, \( q_A' = 2.75 \, \mu C \)
- Charge on disc B, \( q_B' = 8.25 \, \mu C \)