A parallel-plate air capacitor of area A = 12.5 cm2 and plate separation of d = 2.50 mm is charged by a battery to a voltage of 66.0 V. What is the charge on the capacitor?

To find the charge \( Q \) on a parallel-plate capacitor, we can use the formula:

\[
Q = C \cdot V
\]

where:
- \( Q \) is the charge,
- \( C \) is the capacitance, and
- \( V \) is the voltage applied across the plates.

### Step 1: Find the capacitance \( C \)
The capacitance \( C \) of a parallel-plate capacitor is given by:

\[
C = \frac{\varepsilon_0 \cdot A}{d}
\]

where:
- \( \varepsilon_0 = 8.85 \times 10^{-12} \, \text{F/m} \) (the permittivity of free space),
- \( A \) is the area of the plates (in m²),
- \( d \) is the separation between the plates (in meters).

First, convert the given values to SI units:
- \( A = 12.5 \, \text{cm}^2 = 12.5 \times 10^{-4} \, \text{m}^2 \),
- \( d = 2.50 \, \text{mm} = 2.50 \times 10^{-3} \, \text{m} \),
- \( V = 66.0 \, \text{V} \).

Now calculate the capacitance:

\[
C = \frac{(8.85 \times 10^{-12}) \cdot (12.5 \times 10^{-4})}{2.50 \times 10^{-3}} 
\]

\[
C \approx 4.425 \times 10^{-12} \, \text{F}
\]

### Step 2: Calculate the charge \( Q \)
Now, using \( C \) and \( V \) in the formula \( Q = C \cdot V \):

\[
Q = (4.425 \times 10^{-12} \, \text{F}) \cdot (66.0 \, \text{V})
\]

\[
Q \approx 2.92 \times 10^{-10} \, \text{C}
\]

Thus, the charge on the capacitor is approximately \( 2.92 \times 10^{-10} \, \text{C} \).

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Would you like further details or have any questions? 

Here are 5 related questions to consider:
1. How does changing the plate separation affect the capacitance?
2. What happens to the charge if the voltage is doubled?
3. How would the result differ if a dielectric material was introduced between the plates?
4. What is the energy stored in this capacitor?
5. How does the capacitor's charge change if the area of the plates is increased?

**Tip:** Always make sure to convert all given units to SI units when solving physics problems.

Calculate the charge on a parallel-plate air capacitor with an area of 12.5 cm2 and a separation of 2.50 mm, charged to a voltage of 66.0 V. This solution covers capacitance, the formula Q = C * V, and relevant unit conversions. Suitable for Grades 10-12.

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