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An air parallel plate capacitor has capa...

An air parallel plate capacitor has capacity C. When the area and distance between the plates is doubled, the capacitance is `C_(1)`, then `(C_(1))/(C )` is -

A

1

B

2

C

3

D

4

Text Solution

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The correct Answer is:
To solve the problem, we need to analyze how the capacitance of a parallel plate capacitor changes when the area and distance between the plates are modified. ### Step-by-Step Solution: 1. **Understand the Formula for Capacitance**: The capacitance \( C \) of a parallel plate capacitor is given by the formula: \[ C = \frac{A \cdot \epsilon_0}{D} \] where: - \( C \) = capacitance - \( A \) = area of the plates - \( \epsilon_0 \) = permittivity of free space (a constant) - \( D \) = distance between the plates 2. **Identify the Changes**: According to the problem, both the area \( A \) and the distance \( D \) are doubled. Therefore: - New area \( A_1 = 2A \) - New distance \( D_1 = 2D \) 3. **Calculate the New Capacitance**: Substitute the new values into the capacitance formula to find the new capacitance \( C_1 \): \[ C_1 = \frac{A_1 \cdot \epsilon_0}{D_1} = \frac{(2A) \cdot \epsilon_0}{2D} \] 4. **Simplify the Expression**: Simplifying the equation gives: \[ C_1 = \frac{2A \cdot \epsilon_0}{2D} = \frac{A \cdot \epsilon_0}{D} = C \] 5. **Calculate the Ratio of the New Capacitance to the Original Capacitance**: Now, we need to find the ratio \( \frac{C_1}{C} \): \[ \frac{C_1}{C} = \frac{C}{C} = 1 \] ### Final Answer: Thus, the ratio \( \frac{C_1}{C} \) is \( 1 \). ---

To solve the problem, we need to analyze how the capacitance of a parallel plate capacitor changes when the area and distance between the plates are modified. ### Step-by-Step Solution: 1. **Understand the Formula for Capacitance**: The capacitance \( C \) of a parallel plate capacitor is given by the formula: \[ C = \frac{A \cdot \epsilon_0}{D} ...
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