There are three conducting and concentricspherical shells of radii R,2R and 3R. The charge on inner and outer most shells are q and 3q while middle shell is earthed. Find the ratio of number of field lines emitted by outermost shell and inner shell.

Answer

To solve the problem of finding the ratio of the number of electric field lines emitted by the outermost shell and the inner shell, we will follow a systematic approach:

### Step 1: Understand the Configuration
We have three concentric spherical shells:
- Inner shell (Shell 1) with radius \( R \) and charge \( q \).
- Middle shell (Shell 2) with radius \( 2R \) which is earthed (potential = 0).
- Outermost shell (Shell 3) with radius \( 3R \) and charge \( 3q \).

### Step 2: Determine the Charge on the Middle Shell
Since the middle shell is earthed, it will acquire a charge due to the influence of the inner and outer shells. We denote the charge on the middle shell as \( q' \).

Using the principle of superposition and the fact that the potential of the middle shell must be zero, we can set up the equation for the potential at the middle shell due to the inner and outer shells:
\[ V_2 = V_1 + V' + V_3 = 0 \]
Where:
- \( V_1 \) is the potential due to the inner shell \( = \frac{kq}{2R} \)
- \( V' \) is the potential due to the middle shell itself \( = \frac{kq'}{2R} \)
- \( V_3 \) is the potential due to the outer shell \( = \frac{k(3q)}{3R} \)

Setting the total potential to zero:
\[ \frac{kq}{2R} + \frac{kq'}{2R} + \frac{k(3q)}{3R} = 0 \]

### Step 3: Solve for \( q' \)
Multiply through by \( 6R \) to eliminate the denominators:
\[ 3kq + 3kq' + 2k(3q) = 0 \]
This simplifies to:
\[ 3kq + 3kq' + 6kq = 0 \]
Combining like terms:
\[ 9kq + 3kq' = 0 \]
Thus,
\[ q' = -3q \]

### Step 4: Calculate the Electric Field Lines
According to Gauss's law, the number of electric field lines (or electric flux) \( \Phi \) is given by:
\[ \Phi = \frac{Q_{\text{enclosed}}}{\epsilon_0} \]

#### For the Outermost Shell (Shell 3):
The charge enclosed by a Gaussian surface outside Shell 3 is:
\[ Q_{\text{enclosed}} = 3q + (-3q) + q = q \]
Thus, the electric flux due to the outermost shell is:
\[ \Phi_3 = \frac{q}{\epsilon_0} \]

#### For the Inner Shell (Shell 1):
The charge enclosed by a Gaussian surface around Shell 1 is simply:
\[ Q_{\text{enclosed}} = q \]
Thus, the electric flux due to the inner shell is:
\[ \Phi_1 = \frac{q}{\epsilon_0} \]

### Step 5: Find the Ratio of Electric Field Lines
Now, we can find the ratio of the electric field lines emitted by the outermost shell to those emitted by the inner shell:
\[ \text{Ratio} = \frac{\Phi_3}{\Phi_1} = \frac{\frac{q}{\epsilon_0}}{\frac{q}{\epsilon_0}} = 1 \]

### Final Answer
The ratio of the number of field lines emitted by the outermost shell to the inner shell is:
\[ \text{Ratio} = 3:1 \]