A capacitor with air as the dielectric is charged to a potential of `100` volts. If the space between the plates is now filled with a dielectric of dielectric constant `10`, the potential difference between the plates will be

Options

A. `1000` volts
B. `100` volts
C. `10` volts
D. Zero

Explanation

To solve the problem, we need to understand the relationship between capacitance, charge, and voltage in a capacitor. We will follow these steps:

### Step 1: Understand the Initial Conditions
Initially, the capacitor is charged to a potential of 100 volts with air as the dielectric. The capacitance of a capacitor with air as the dielectric can be denoted as \( C_0 \).

### Step 2: Calculate the Initial Charge
The charge \( Q \) on the capacitor can be calculated using the formula:
\[ Q = C_0 \cdot V_0 \]
where \( V_0 = 100 \, \text{volts} \).

### Step 3: Introduce the Dielectric
When the space between the plates is filled with a dielectric of dielectric constant \( K = 10 \), the new capacitance \( C \) becomes:
\[ C = K \cdot C_0 = 10 \cdot C_0 \]

### Step 4: Charge Conservation
The charge on the capacitor remains constant when the dielectric is introduced, so we have:
\[ Q = Q' = C_0 \cdot V_0 = C \cdot V' \]
where \( V' \) is the new potential difference across the capacitor.

### Step 5: Substitute for Charge
Substituting the expressions for charge, we get:
\[ C_0 \cdot V_0 = (10 \cdot C_0) \cdot V' \]

### Step 6: Simplify the Equation
We can cancel \( C_0 \) from both sides (assuming \( C_0 \neq 0 \)):
\[ V_0 = 10 \cdot V' \]

### Step 7: Solve for the New Voltage
Now, substituting \( V_0 = 100 \, \text{volts} \):
\[ 100 = 10 \cdot V' \]
\[ V' = \frac{100}{10} = 10 \, \text{volts} \]

### Conclusion
The potential difference between the plates after filling the space with the dielectric is \( 10 \, \text{volts} \).

### Final Answer
The potential difference between the plates will be **10 volts**.

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