The capacitance of a parallel plate capacitor is `10muF`. When a dielectric plate is introduced in between the plates, its potential becomes 1/4th of its original value. What is the value of the dielectric constant of the plate introduced?

Options

A. 4
B. 40
C. 2.5
D. none of the above

Explanation

To solve the problem step by step, we will use the relationship between capacitance, charge, and potential difference in a capacitor, along with the effect of a dielectric material.

### Step-by-Step Solution:

1. **Identify Given Values**:
- The capacitance of the capacitor, \( C = 10 \mu F = 10 \times 10^{-6} F \).
- The potential after introducing the dielectric becomes \( \frac{1}{4} \) of its original value.

2. **Understand the Relationship**:
- The potential \( V \) across a capacitor is given by the formula:
\[ V = \frac{Q}{C} \]
- When a dielectric with dielectric constant \( k \) is introduced, the new capacitance \( C' \) becomes:
\[ C' = kC \]
- The new potential \( V' \) can be expressed as:
\[ V' = \frac{Q}{C'} = \frac{Q}{kC} \]

3. **Express the New Potential in Terms of the Original Potential**:
- From the problem, we know that:
\[ V' = \frac{1}{4} V \]
- Substituting the expression for \( V' \):
\[ \frac{Q}{kC} = \frac{1}{4} \cdot \frac{Q}{C} \]

4. **Cancel Out Common Terms**:
- Since \( Q \) and \( C \) are common in both sides (and assuming \( Q \neq 0 \) and \( C \neq 0 \)), we can simplify:
\[ \frac{1}{k} = \frac{1}{4} \]

5. **Solve for the Dielectric Constant \( k \)**:
- Rearranging gives:
\[ k = 4 \]

6. **Conclusion**:
- The value of the dielectric constant of the plate introduced is \( k = 4 \).

### Final Answer:
The dielectric constant \( k \) is \( 4 \).

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