In the above problem, if the battery is disconnected before inserting the dielectric, then difference will be

Options

A. 0.5 V
B. 2 V
C. 4 V
D. 32 V

Answer

To solve the problem step by step, we need to analyze the situation when a dielectric is inserted into a capacitor after the battery has been disconnected.

### Step-by-Step Solution:

1. **Understand the Initial Setup:**
- A capacitor is connected to a battery (4 volts in this case) which charges the capacitor to a certain charge \( Q \).
- When the battery is disconnected, the charge \( Q \) on the capacitor remains constant because there is no path for charge to flow.

**Hint:** Recall that when a battery is connected to a capacitor, it charges up to a voltage equal to the battery voltage.

2. **Insert the Dielectric:**
- When a dielectric material is inserted between the plates of the capacitor, the capacitance \( C \) of the capacitor increases. The new capacitance with the dielectric can be expressed as \( C' = kC \), where \( k \) is the dielectric constant.

**Hint:** Remember that inserting a dielectric increases the capacitance of the capacitor.

3. **Relate Charge, Voltage, and Capacitance:**
- The relationship between charge \( Q \), capacitance \( C \), and voltage \( V \) is given by the formula:
\[ Q = CV \]
- Since \( Q \) is constant (because the battery is disconnected), we can express the new voltage \( V' \) after inserting the dielectric as:
\[ Q = C'V' \quad \text{and} \quad Q = CV \]
- Therefore, we can write:
\[ CV = C'V' \]

**Hint:** Use the relationship \( Q = CV \) to express how the voltage changes when the capacitance changes.

4. **Substituting for Capacitance:**
- Substitute \( C' = kC \) into the equation:
\[ CV = kCV' \]
- Dividing both sides by \( C \) (assuming \( C \neq 0 \)):
\[ V = kV' \]
- Rearranging gives:
\[ V' = \frac{V}{k} \]

**Hint:** Isolate \( V' \) to find how the voltage changes with the dielectric constant.

5. **Calculate the New Voltage:**
- If the dielectric constant \( k = 8 \) (as suggested in the video transcript), then:
\[ V' = \frac{V}{8} \]
- Given that the initial voltage \( V = 4 \) volts, we find:
\[ V' = \frac{4}{8} = 0.5 \text{ volts} \]

**Hint:** Substitute the known values to find the new voltage across the capacitor.

### Final Answer:
The new voltage across the capacitor after inserting the dielectric is \( 0.5 \) volts.