https://file.aiquickdraw.com/mathbot/file/17304728951548wvgzcz5.jpg

Let's solve each part step-by-step. Here's the information provided:

- **Area of plates, \( A = 2.90 \, \text{cm}^2 \)**
- **Separation distance, \( d = 1.00 \, \text{mm} \)**
- **Voltage, \( V = 15.0 \, \text{V} \)**

We'll go through each part:

### (a) Find the capacitance of the capacitor

The capacitance \( C \) of a parallel-plate capacitor is given by the formula:
\[
C = \frac{\varepsilon_0 A}{d}
\]
where:
- \( \varepsilon_0 \) is the permittivity of free space (\( \varepsilon_0 = 8.85 \times 10^{-12} \, \text{F/m} \)),
- \( A \) is the area of the plates (in square meters),
- \( d \) is the separation between the plates (in meters).

#### Step 1: Convert units
1. **Convert \( A = 2.90 \, \text{cm}^2 \) to square meters:**
   \[
   A = 2.90 \times 10^{-4} \, \text{m}^2
   \]
2. **Convert \( d = 1.00 \, \text{mm} \) to meters:**
   \[
   d = 1.00 \times 10^{-3} \, \text{m}
   \]

#### Step 2: Calculate capacitance
\[
C = \frac{(8.85 \times 10^{-12}) \times (2.90 \times 10^{-4})}{1.00 \times 10^{-3}}
\]
\[
C \approx 2.57 \times 10^{-12} \, \text{F} = 2.57 \, \text{pF}
\]

### (b) Find the charge on the capacitor

The charge \( Q \) on a capacitor is given by:
\[
Q = CV
\]
where:
- \( C \) is the capacitance we found in part (a),
- \( V = 15.0 \, \text{V} \).

Substitute the values:
\[
Q = (2.57 \times 10^{-12} \, \text{F}) \times (15.0 \, \text{V})
\]
\[
Q \approx 3.86 \times 10^{-11} \, \text{C} = 38.6 \, \text{pC}
\]

### (c) Find the magnitude of the electric field between the plates

The electric field \( E \) between the plates of a parallel-plate capacitor is given by:
\[
E = \frac{V}{d}
\]

Substitute the values:
\[
E = \frac{15.0 \, \text{V}}{1.00 \times 10^{-3} \, \text{m}}
\]
\[
E = 1.5 \times 10^4 \, \text{V/m}
\]

### Summary of Answers:
1. **Capacitance, \( C \approx 2.57 \, \text{pF} \)**
2. **Charge, \( Q \approx 38.6 \, \text{pC} \)**
3. **Electric Field, \( E = 1.5 \times 10^4 \, \text{V/m} \)**

Would you like further details on any part?

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Here are some related questions you may want to explore:
1. How does the capacitance change if the separation between the plates is doubled?
2. What effect does inserting a dielectric material between the plates have on the capacitance?
3. How would the charge change if the voltage were reduced to 7.5 V?
4. What would happen to the electric field if the area of the plates was increased?
5. How is the energy stored in the capacitor related to its capacitance and voltage?

**Tip:** Always check the units in physics problems to avoid errors and ensure accurate calculations.

An air-filled parallel-plate capacitor has plates of area 2.90 cm² separated by 1.00 mm. The capacitor is connected to a 15.0-V battery.
(a) Find the value of its capacitance in pF.
(b) What is the charge on the capacitor in pC?
(c) What is the magnitude of the uniform electric field between the plates in V/m?

Solve a physics problem involving an air-filled parallel-plate capacitor with plates of 2.90 cm² area separated by 1.00 mm, connected to a 15.0-V battery. Learn to calculate the capacitance, charge on the capacitor, and the magnitude of the uniform electric field between the plates. Perfect for college-level introductory physics.

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