10. (II) An electric field of 8.50 × 10³ V/m is desired between two parallel plates, each of area 45.0 cm² and separated by 2.45 mm of air. What charge must be on each plate?
10. (II) An electric field of 8.50 × 10³ V/m is desired between two parallel plates, each of area 45.0 cm² and separated by 2.45 mm of air. What charge must be on each plate?
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
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Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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please type your solution so that it is easy for to read I have bad eyesight please and thank you
![**Problem 40: Determining Charge on Parallel Plates**
*Description:*
Given an electric field of \(8.50 \times 10^3 \text{ V/m}\) between two parallel plates, each having an area of \(45.0 \text{ cm}^2\) and separated by \(2.45 \text{ mm}\) of air, we need to calculate the charge on each plate.
**Solution Steps:**
1. **Convert units of area and separation distance:**
- Area of each plate (\(A\)):
\[
A = 45.0 \text{ cm}^2 = 45.0 \times 10^{-4} \text{ m}^2 = 4.50 \times 10^{-3} \text{ m}^2
\]
- Separation distance (\(d\)):
\[
d = 2.45 \text{ mm} = 2.45 \times 10^{-3} \text{ m}
\]
2. **Determine the electric field (\(E\)):**
Given:
\[
E = 8.50 \times 10^3 \text{ V/m}
\]
3. **Use the relationship between the electric field and voltage (\(E = V/d\)):**
Rearrange to solve for voltage (\(V\)):
\[
V = E \times d = 8.50 \times 10^3 \text{ V/m} \times 2.45 \times 10^{-3} \text{ m}
\]
\[
V = 20.825 \text{ V}
\]
4. **Determine the capacitance (\(C\)) of the parallel plates:**
The capacitance is given by:
\[
C = \epsilon_0 \frac{A}{d}
\]
where \( \epsilon_0 \) (the permittivity of free space) is \(8.85 \times 10^{-12} \text{ F/m}\).
\[
C = 8.85 \times 10^{-12} \frac{4.50 \times 10^{-3}}{2.45 \times 10^{-3}}
\]
\[
C = 1.62 \times 10^{-](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F100ca580-01b3-4065-a77f-0a16a29f1100%2Fc601ffcb-832f-452f-bbd2-5c1c6ec9d1da%2Fp5tshao_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem 40: Determining Charge on Parallel Plates**
*Description:*
Given an electric field of \(8.50 \times 10^3 \text{ V/m}\) between two parallel plates, each having an area of \(45.0 \text{ cm}^2\) and separated by \(2.45 \text{ mm}\) of air, we need to calculate the charge on each plate.
**Solution Steps:**
1. **Convert units of area and separation distance:**
- Area of each plate (\(A\)):
\[
A = 45.0 \text{ cm}^2 = 45.0 \times 10^{-4} \text{ m}^2 = 4.50 \times 10^{-3} \text{ m}^2
\]
- Separation distance (\(d\)):
\[
d = 2.45 \text{ mm} = 2.45 \times 10^{-3} \text{ m}
\]
2. **Determine the electric field (\(E\)):**
Given:
\[
E = 8.50 \times 10^3 \text{ V/m}
\]
3. **Use the relationship between the electric field and voltage (\(E = V/d\)):**
Rearrange to solve for voltage (\(V\)):
\[
V = E \times d = 8.50 \times 10^3 \text{ V/m} \times 2.45 \times 10^{-3} \text{ m}
\]
\[
V = 20.825 \text{ V}
\]
4. **Determine the capacitance (\(C\)) of the parallel plates:**
The capacitance is given by:
\[
C = \epsilon_0 \frac{A}{d}
\]
where \( \epsilon_0 \) (the permittivity of free space) is \(8.85 \times 10^{-12} \text{ F/m}\).
\[
C = 8.85 \times 10^{-12} \frac{4.50 \times 10^{-3}}{2.45 \times 10^{-3}}
\]
\[
C = 1.62 \times 10^{-
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