A uniform electric field of magnitude 2.7 x 10* N/C is perpendicular to a square sheet with sides 2.5 m long. What is the electric flux through the sheet? Hint Electric flux is PE N.m²/C.
A uniform electric field of magnitude 2.7 x 10* N/C is perpendicular to a square sheet with sides 2.5 m long. What is the electric flux through the sheet? Hint Electric flux is PE N.m²/C.
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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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![**Problem: Electric Flux Through a Square Sheet**
A uniform electric field of magnitude \(2.7 \times 10^4 \, \text{N/C}\) is perpendicular to a square sheet with sides \(2.5 \, \text{m}\) long. What is the electric flux through the sheet?
---
**Solution:**
**Electric Flux Formula:**
The electric flux \(\Phi_E\) through a surface is given by:
\[
\Phi_E = E \cdot A \cdot \cos(\theta)
\]
where:
- \(E\) is the magnitude of the electric field.
- \(A\) is the area of the square.
- \(\theta\) is the angle between the electric field and the normal to the surface.
**Step 1: Calculate the Area (\(A\)) of the Square Sheet**
The area of a square is given by:
\[
A = \text{side}^2 = (2.5 \, \text{m})^2 = 6.25 \, \text{m}^2
\]
**Step 2: Determine the Angle (\(\theta\))**
Since the electric field is perpendicular to the sheet, \(\theta = 0^\circ\).
**Step 3: Calculate the Electric Flux (\(\Phi_E\))**
Substitute the values into the electric flux formula:
\[
\Phi_E = (2.7 \times 10^4 \, \text{N/C}) \cdot (6.25 \, \text{m}^2) \cdot \cos(0^\circ)
\]
Since \(\cos(0^\circ) = 1\):
\[
\Phi_E = (2.7 \times 10^4 \, \text{N/C}) \cdot (6.25 \, \text{m}^2) = 1.6875 \times 10^5 \, \text{N} \cdot \text{m}^2/\text{C}
\]
Therefore, the electric flux through the sheet is:
\[
\Phi_E = 1.6875 \times 10^5 \, \text{N} \cdot \text{m}^2/\text{C}
\]
**Additional Features:**
- **Hint Button:** Provides guidance on formulae](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc0edab40-1f13-4f88-959e-31b8927d437a%2Fe39c14a7-eb16-4bcd-961a-bb5e0823625c%2F6fweok_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem: Electric Flux Through a Square Sheet**
A uniform electric field of magnitude \(2.7 \times 10^4 \, \text{N/C}\) is perpendicular to a square sheet with sides \(2.5 \, \text{m}\) long. What is the electric flux through the sheet?
---
**Solution:**
**Electric Flux Formula:**
The electric flux \(\Phi_E\) through a surface is given by:
\[
\Phi_E = E \cdot A \cdot \cos(\theta)
\]
where:
- \(E\) is the magnitude of the electric field.
- \(A\) is the area of the square.
- \(\theta\) is the angle between the electric field and the normal to the surface.
**Step 1: Calculate the Area (\(A\)) of the Square Sheet**
The area of a square is given by:
\[
A = \text{side}^2 = (2.5 \, \text{m})^2 = 6.25 \, \text{m}^2
\]
**Step 2: Determine the Angle (\(\theta\))**
Since the electric field is perpendicular to the sheet, \(\theta = 0^\circ\).
**Step 3: Calculate the Electric Flux (\(\Phi_E\))**
Substitute the values into the electric flux formula:
\[
\Phi_E = (2.7 \times 10^4 \, \text{N/C}) \cdot (6.25 \, \text{m}^2) \cdot \cos(0^\circ)
\]
Since \(\cos(0^\circ) = 1\):
\[
\Phi_E = (2.7 \times 10^4 \, \text{N/C}) \cdot (6.25 \, \text{m}^2) = 1.6875 \times 10^5 \, \text{N} \cdot \text{m}^2/\text{C}
\]
Therefore, the electric flux through the sheet is:
\[
\Phi_E = 1.6875 \times 10^5 \, \text{N} \cdot \text{m}^2/\text{C}
\]
**Additional Features:**
- **Hint Button:** Provides guidance on formulae
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