Two identical condensers M and N are connected in series with a battery. The space between the plates of 'M' is completely filled with dielectric medium of dielectric constant 8 and a copper plate of thickness d/ 2 is introduced between the plates of N. (d = distance of separation of the plates). Then the potentials of M and N are respectively.

Options

A. `1 :4`
B. `4 :1`
C. `3 :8`
D. `1 :6`

Answer

To solve the problem, we need to analyze the two capacitors M and N connected in series and calculate their respective potentials. Here’s a step-by-step solution:

### Step 1: Understand the configuration of capacitors
We have two identical capacitors, M and N. Capacitor M is filled with a dielectric medium of dielectric constant \( k = 8 \). Capacitor N has a copper plate of thickness \( \frac{d}{2} \) inserted between its plates, where \( d \) is the distance between the plates.

### Step 2: Calculate the capacitance of capacitor M
The capacitance of a capacitor filled with a dielectric is given by:
\[ C_m = k \cdot C_0 \]
where \( C_0 \) is the original capacitance without the dielectric. Since \( k = 8 \), we have:
\[ C_m = 8C_0 \]

### Step 3: Calculate the capacitance of capacitor N
For capacitor N, the effective distance between the plates after inserting the copper plate is \( d - \frac{d}{2} = \frac{d}{2} \). The capacitance of a capacitor is inversely proportional to the distance between the plates:
\[ C_n = \frac{\epsilon_0 A}{\frac{d}{2}} = \frac{2\epsilon_0 A}{d} = 2C_0 \]

### Step 4: Relate the charges and potentials in series
Since capacitors M and N are connected in series, the charge \( Q \) on both capacitors is the same. The relationship between charge, capacitance, and potential is given by:
\[ Q = C_m V_m = C_n V_n \]
From this, we can express the potentials:
\[ V_m = \frac{Q}{C_m} \quad \text{and} \quad V_n = \frac{Q}{C_n} \]

### Step 5: Find the ratio of potentials
Using the expressions for \( V_m \) and \( V_n \), we can find the ratio:
\[ \frac{V_m}{V_n} = \frac{C_n}{C_m} \]
Substituting the values of capacitance:
\[ \frac{V_m}{V_n} = \frac{2C_0}{8C_0} = \frac{2}{8} = \frac{1}{4} \]

### Step 6: Conclusion
Thus, the potentials of M and N are in the ratio:
\[ V_m : V_n = 1 : 4 \]