A parallel plate capacitor of capacitance `10muF` is connected to a cell of emf 10 V and is fully charged. Now a dielectric slab (k = 3) of thickness equal to the gap between the plates is completely filled in the gap, keeping the cell connected. During the filling process,

Options

A. the increase in charge on the capacitor is `200 muC` .
B. the heat produced is zero
C. energy supplied by the cell = increase in stored potential energy + work done on the person who is filling the dielectric slab.
D. energy supplied by the cell = increase in stored potential energy + work done on the person who is filling the dielectric slab + heat produced.

Explanation

To solve the problem step by step, we will analyze the situation involving a parallel plate capacitor, the effects of inserting a dielectric slab, and the associated energy changes.

### Step 1: Calculate the Initial Charge on the Capacitor
The initial charge \( Q_{\text{initial}} \) on the capacitor can be calculated using the formula:
\[ Q_{\text{initial}} = C \times V \]
Where:
- \( C = 10 \, \mu F = 10 \times 10^{-6} \, F \)
- \( V = 10 \, V \)

Substituting the values:
\[ Q_{\text{initial}} = 10 \times 10^{-6} \times 10 = 100 \, \mu C \]

### Step 2: Calculate the Final Charge on the Capacitor after Inserting the Dielectric
When a dielectric slab with dielectric constant \( k = 3 \) is inserted, the capacitance increases. The new capacitance \( C' \) can be calculated as:
\[ C' = k \times C = 3 \times 10 \, \mu F = 30 \, \mu F \]

Now, we can find the final charge \( Q_{\text{final}} \) using the new capacitance:
\[ Q_{\text{final}} = C' \times V = 30 \times 10^{-6} \times 10 = 300 \, \mu C \]

### Step 3: Calculate the Increase in Charge
The increase in charge \( \Delta Q \) can be calculated as:
\[ \Delta Q = Q_{\text{final}} - Q_{\text{initial}} = 300 \, \mu C - 100 \, \mu C = 200 \, \mu C \]

### Step 4: Analyze Heat Production
During the filling process of the dielectric slab, since the slab is inserted while the capacitor is connected to the cell, there is no heat produced. This is because the process is quasi-static, and the energy supplied by the cell is entirely used to increase the potential energy stored in the capacitor.

### Step 5: Energy Supplied by the Cell
The energy supplied by the cell can be expressed as:
\[ \text{Energy supplied} = \text{Increase in stored potential energy} + \text{Work done on the person filling the slab} \]

### Step 6: Conclusion on Options
1. The increase in charge on the capacitor is \( 200 \, \mu C \) (True).
2. Heat produced is \( 0 \) (True).
3. Energy supplied by the cell is equal to energy stored as potential energy plus work done on the person filling the dielectric slab (True).
4. Energy supplied by the cell should be equal to increase in stored potential energy plus work done on the person filling the dielectric slab plus heat produced (False, because heat produced is zero).

### Final Answers
The correct options are 1, 2, and 3.

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