Two parallel plates 11 cm on a side are given equal and opposite charges of magnitude 5.2×10-⁹ C. The plates are 1.5 mm apart. What is the electric field at the center of the region between the plates? ×104 N/C E =
Two parallel plates 11 cm on a side are given equal and opposite charges of magnitude 5.2×10-⁹ C. The plates are 1.5 mm apart. What is the electric field at the center of the region between the plates? ×104 N/C E =
Related questions
Question
![### Problem Statement: Electric Field Between Parallel Plates
**Introduction:**
Two parallel plates, each with a side length of 11 cm, are given equal and opposite charges of magnitude \( 5.2 \times 10^{-9} \) C. The distance between the plates is 1.5 mm. We need to determine the electric field at the center of the region between the plates.
**Question:**
What is the electric field at the center of the region between the plates?
**Equation:**
\[ E = \_\_\_\_ \times 10^4 \, \text{N/C} \]
### Explanation:
Given:
- Side length of plates: 11 cm
- Charge on plates: \( 5.2 \times 10^{-9} \) C
- Distance between plates: 1.5 mm
To find:
- Electric field (E) at the center of the region between the plates
### Solution Approach:
1. **Calculate the area (A) of each plate:**
Given side length \( l = 11 \, \text{cm} = 0.11 \, \text{m} \)
\[ A = l^2 = (0.11 \, \text{m})^2 = 0.0121 \, \text{m}^2 \]
2. **Electric field (E) created between parallel plates:**
The electric field between two parallel plates is given by:
\[ E = \frac{\sigma}{\epsilon_0} \]
where:
\[ \sigma = \frac{Q}{A} \, \text{(surface charge density)} \]
\[ \epsilon_0 = 8.85 \times 10^{-12} \, \text{F/m} \, \text{(permittivity of free space)} \]
Substituting the given values:
\[ \sigma = \frac{Q}{A} = \frac{5.2 \times 10^{-9}}{0.0121} = 4.3 \times 10^{-7} \, \text{C/m}^2 \]
Therefore,
\[ E = \frac{4.3 \times 10^{-7}}{8.85 \times 10^{-12}} = 4.86 \times 10](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9bd1217a-91ba-44de-b552-38ea0e6629c6%2Fb5d72e82-d716-432c-9b79-6210e1614652%2Fbwz47sf_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement: Electric Field Between Parallel Plates
**Introduction:**
Two parallel plates, each with a side length of 11 cm, are given equal and opposite charges of magnitude \( 5.2 \times 10^{-9} \) C. The distance between the plates is 1.5 mm. We need to determine the electric field at the center of the region between the plates.
**Question:**
What is the electric field at the center of the region between the plates?
**Equation:**
\[ E = \_\_\_\_ \times 10^4 \, \text{N/C} \]
### Explanation:
Given:
- Side length of plates: 11 cm
- Charge on plates: \( 5.2 \times 10^{-9} \) C
- Distance between plates: 1.5 mm
To find:
- Electric field (E) at the center of the region between the plates
### Solution Approach:
1. **Calculate the area (A) of each plate:**
Given side length \( l = 11 \, \text{cm} = 0.11 \, \text{m} \)
\[ A = l^2 = (0.11 \, \text{m})^2 = 0.0121 \, \text{m}^2 \]
2. **Electric field (E) created between parallel plates:**
The electric field between two parallel plates is given by:
\[ E = \frac{\sigma}{\epsilon_0} \]
where:
\[ \sigma = \frac{Q}{A} \, \text{(surface charge density)} \]
\[ \epsilon_0 = 8.85 \times 10^{-12} \, \text{F/m} \, \text{(permittivity of free space)} \]
Substituting the given values:
\[ \sigma = \frac{Q}{A} = \frac{5.2 \times 10^{-9}}{0.0121} = 4.3 \times 10^{-7} \, \text{C/m}^2 \]
Therefore,
\[ E = \frac{4.3 \times 10^{-7}}{8.85 \times 10^{-12}} = 4.86 \times 10
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 3 steps with 2 images
