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The plate area of a parallel-plate capac...

The plate area of a parallel-plate capacitors is `10cm^(2)` and its capacitance is `2pF`. The separation between the plates of the capacitors is close to (in mm).
(permittivity of vacuum, `epsi_(0)=(80)/(9)xx10^(-12)m^(-3)kg^(-1)s^(4)A^(2),1pF=10^(-12)F` )

A

2.2

B

4.4

C

8.8

D

17.6

Text Solution

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The correct Answer is:
To find the separation between the plates of a parallel-plate capacitor, we can use the formula for capacitance: \[ C = \frac{A \epsilon_0}{d} \] Where: - \(C\) is the capacitance, - \(A\) is the area of the plates, - \(\epsilon_0\) is the permittivity of free space, - \(d\) is the separation between the plates. ### Step 1: Convert the given values to SI units 1. **Area**: The area \(A\) is given as \(10 \, \text{cm}^2\). We need to convert this to square meters: \[ A = 10 \, \text{cm}^2 = 10 \times 10^{-4} \, \text{m}^2 = 1 \times 10^{-3} \, \text{m}^2 \] 2. **Capacitance**: The capacitance \(C\) is given as \(2 \, \text{pF}\). We need to convert this to farads: \[ C = 2 \, \text{pF} = 2 \times 10^{-12} \, \text{F} \] ### Step 2: Use the formula to find \(d\) Rearranging the capacitance formula to solve for \(d\): \[ d = \frac{A \epsilon_0}{C} \] ### Step 3: Substitute the values into the equation 1. **Permittivity of free space**: The value of \(\epsilon_0\) is given as: \[ \epsilon_0 = \frac{80}{9} \times 10^{-12} \, \text{m}^{-3}\text{kg}^{-1}\text{s}^4\text{A}^2 \] 2. **Substituting the values**: \[ d = \frac{(1 \times 10^{-3} \, \text{m}^2) \left(\frac{80}{9} \times 10^{-12} \, \text{F/m}\right)}{2 \times 10^{-12} \, \text{F}} \] ### Step 4: Calculate \(d\) Calculating the numerator: \[ d = \frac{(1 \times 10^{-3}) \left(\frac{80}{9} \times 10^{-12}\right)}{2 \times 10^{-12}} \] \[ = \frac{(80 \times 10^{-15})}{(2 \times 9)} \, \text{m} \] \[ = \frac{80}{18} \times 10^{-3} \, \text{m} \] \[ = \frac{40}{9} \times 10^{-3} \, \text{m} \] \[ \approx 4.44 \times 10^{-3} \, \text{m} = 4.44 \, \text{mm} \] ### Final Answer The separation between the plates of the capacitor is approximately **4.44 mm**. ---

To find the separation between the plates of a parallel-plate capacitor, we can use the formula for capacitance: \[ C = \frac{A \epsilon_0}{d} \] Where: - \(C\) is the capacitance, ...
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