40. (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?
40. (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
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ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
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 pleae and thank you
![### Electrostatics Problem: Determining the Charge on Parallel Plates
#### Problem Statement:
An electric field of \(8.50 \times 10^3 \) V/m is desired between two parallel plates, each of an area of \(45.0 \text{ cm}^2\), and separated by \(2.45 \text{ mm}\) of air. What charge must be on each plate?
#### Detailed Explanation:
**Given Data:**
- Electric field (\(E\)): \(8.50 \times 10^3 \) V/m
- Area of each plate (\(A\)): \(45.0 \text{ cm}^2\)
- Distance between the plates (\(d\)): \(2.45 \text{ mm}\)
**Required:**
The charge (\(Q\)) on each plate.
**Solution Strategy:**
1. **Convert units:**
- Area (\(A\)) from cm\(^2\) to m\(^2\):
\[
45.0 \text{ cm}^2 = 45.0 \times 10^{-4} \text{ m}^2 = 4.50 \times 10^{-3} \text{ m}^2
\]
- Distance (\(d\)) from mm to meters:
\[
2.45 \text{ mm} = 2.45 \times 10^{-3} \text{ m}
\]
2. **Use the formula relating electric field (\(E\)) and charge (\(Q\)) for parallel plates:**
\[
E = \frac{V}{d} \quad \text{where} \quad V = \frac{Qd}{\epsilon_0 A}
\]
Here, \(\epsilon_0\) is the permittivity of free space \(( \epsilon_0 = 8.85 \times 10^{-12} \text{ C}^2/\text{N m}^2)\).
3. **Rearrange the equation to solve for \(Q\):**
\[
E = \frac{Q}{\epsilon_0 A}
\]
\[
Q = E \times \epsilon_0 \times A
\]
4. **Substitute the known values:**
\[](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8c08191f-4610-4f17-af7c-727bd3de0813%2F7ff640e1-0b99-42ab-aff7-ed16e0e8ba5d%2Fpf2u0pc_processed.png&w=3840&q=75)
Transcribed Image Text:### Electrostatics Problem: Determining the Charge on Parallel Plates
#### Problem Statement:
An electric field of \(8.50 \times 10^3 \) V/m is desired between two parallel plates, each of an area of \(45.0 \text{ cm}^2\), and separated by \(2.45 \text{ mm}\) of air. What charge must be on each plate?
#### Detailed Explanation:
**Given Data:**
- Electric field (\(E\)): \(8.50 \times 10^3 \) V/m
- Area of each plate (\(A\)): \(45.0 \text{ cm}^2\)
- Distance between the plates (\(d\)): \(2.45 \text{ mm}\)
**Required:**
The charge (\(Q\)) on each plate.
**Solution Strategy:**
1. **Convert units:**
- Area (\(A\)) from cm\(^2\) to m\(^2\):
\[
45.0 \text{ cm}^2 = 45.0 \times 10^{-4} \text{ m}^2 = 4.50 \times 10^{-3} \text{ m}^2
\]
- Distance (\(d\)) from mm to meters:
\[
2.45 \text{ mm} = 2.45 \times 10^{-3} \text{ m}
\]
2. **Use the formula relating electric field (\(E\)) and charge (\(Q\)) for parallel plates:**
\[
E = \frac{V}{d} \quad \text{where} \quad V = \frac{Qd}{\epsilon_0 A}
\]
Here, \(\epsilon_0\) is the permittivity of free space \(( \epsilon_0 = 8.85 \times 10^{-12} \text{ C}^2/\text{N m}^2)\).
3. **Rearrange the equation to solve for \(Q\):**
\[
E = \frac{Q}{\epsilon_0 A}
\]
\[
Q = E \times \epsilon_0 \times A
\]
4. **Substitute the known values:**
\[
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