### Analyzing Electric Flux Through a Closed Triangular Box#### Problem StatementConsider a closed triangular box resting within a horizontal electric field of magnitude \(E = 6.22 \times 10^4 \, \text{N/C}\) as shown in the figure below.#### Diagram DescriptionThe diagram depicts a triangular box placed in a horizontal electric field. The specified dimensions of the box are:- Height: 10.0 cm- Base: 30.0 cm- Angle between the slanted surface and horizontal base: 60.0 degreesThe electric field \( \vec{E} \) is directed horizontally, depicted with orange arrows pointing from left to right.#### Questions(a) Calculate the electric flux through the vertical rectangular surface of the box.\[ \text{Electric Flux} (kN \cdot m^2/C) \](b) Calculate the electric flux through the slanted surface of the box.\[ \text{Electric Flux} (kN \cdot m^2/C) \](c) Calculate the electric flux through the entire surface of the box.\[ \text{Electric Flux} (kN \cdot m^2/C) \]Ensure to use the principles of electric flux calculation, considering the orientation of each surface with respect to the electric field direction.

Consider a closed triangular box resting within a horizontal electric field of magnitude E = 6.22 x 10¹ N/C as shown in the figure below. 30.0 cm 60.0⁰ 10.0 cm (a) Calculate the electric flux through the vertical rectangular surface of the box. kNm²/C (b) Calculate the electric flux through the slanted surface of the box. kN. m²/C (c) Calculate the electric flux through the entire surface of the box. kN. m²/c

Consider a closed triangular box resting within a horizontal electric field of magnitude E = 6.22 x 10¹ N/C as shown in the figure below. 30.0 cm 60.0⁰ 10.0 cm (a) Calculate the electric flux through the vertical rectangular surface of the box. kNm²/C (b) Calculate the electric flux through the slanted surface of the box. kN. m²/C (c) Calculate the electric flux through the entire surface of the box. kN. m²/c

College Physics

11th Edition

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)...

See similar textbooks

Related questions

Q: 60.0° 10.0 cm 30.0 cm 1 (a) Calculate the electric flux through the vertical rectangular surface of…

Q: Q39. Electric flux through a surface is maximum when O Angle of inclination between the field and…

A: we know that electric flux through the surface is given by φ = φ = E dA cosθ hence flux will be…

Q: Consider a closed triangular box resting within a horizontal electric field of magnitude E = 6.52 x…

Q: A uniform electric field of magnitude 2.7 x 10 N/C is perpendicular to a square sheet with sides 2.5…

A: Magnitude of electric field, E = 2.7 × 10⁴ N/C Length, L = 2.5 m We know, The area will be: A = L²…

Q: A flat circle of radius 31.25 cm is rotated through a uniform horizontal electric field of 7.5 N/C.…

A: The expression for the electric flux is written as, Here, E, A and θ represent the electric field,…

Q: Consider a closed triangular box resting within a horizontal electric field of magnitude E = 7.10 x…

A: Given E = 7.10×104 N/C

Q: Consider a closed triangular box resting within a horizontal electric field of magnitude E = 6.58 x…

A: E = 6.58 x 104 N/C

Q: Suppose three charges: 91 = 4.00 nC, q2 = – 4.00 nC, q3=2.40 nC are enclosed by five surfaces as…

A: Given that, First charge q1=4.00 nC Second charge q2=-4.00 nC Third charge q3=2.40 nC

Q: A closed surface is in the form of a block with length c, width b and height a. In this room there…

A: The objective of the question is to find the total electric flux over the closed surface and the…

Q: Consider a closed triangular box resting within a horizontal electric field of magnitude E = 6.08 x…

Q: An electric field of 5.4 N/C passes through an area of 3.4 cm2. Calculate the flux in units of 10^-4…

A: Given data: Electric field (E) = 5.4 N/C Area (A) = 3.4 cm2 = 3.4×10-4 m2 Angle between electric…

Q: 15.7c Consider a closed triangular box resting within a horizontal electric field of magnitude E =…

Q: Consider a closed triangular box resting within a horizontal electric field of magnitude E = 7.70 x…

Q: A rectangular surface extends from x = 0 to x = 50.0 cm and from y = 0 to y = 42.0 cm in the xy…

Q: A closed surface is in the form of a block with length c, width b and height a. In this room there…

A: The objective of the question is to find the total electric flux over the closed surface and the…

Q: Consider a closed triangular box resting within a horizontal electric field of magnitude E = 6.62 x…

Q: An electric field with a magnitude of 3.40 kN/C is applied along the x axis. (a) Calculate the…

A: Given: The magnitude of the electric field is 3.4 kN/C along the x-direction. The cross-section…

Q: Two thin, square parallel plates 10 cm on a side are given equal and opposite charges of magnitude…

Q: Consider a closed triangular box resting within a horizontal electric field of magnitude E = 8.72 x…

Q: Part A What is the magnitude of the electric flux through the rectangle if E = (150i – 250j) N/C?…

Q: Consider a closed triangular box resting within a horizontal electric field of magnitude E = 7.32 x…

Concept explainers

Electric Charges and Fields

One of the fundamental properties is the electromagnetic property. Charge cannot be destroyed by any process and this contributes formally to the law of charge conservation.

Question

100%

### Analyzing Electric Flux Through a Closed Triangular Box

#### Problem Statement

Consider a closed triangular box resting within a horizontal electric field of magnitude \(E = 6.22 \times 10^4 \, \text{N/C}\) as shown in the figure below.

#### Diagram Description

The diagram depicts a triangular box placed in a horizontal electric field. The specified dimensions of the box are:
- Height: 10.0 cm
- Base: 30.0 cm
- Angle between the slanted surface and horizontal base: 60.0 degrees

The electric field \( \vec{E} \) is directed horizontally, depicted with orange arrows pointing from left to right.

#### Questions

(a) Calculate the electric flux through the vertical rectangular surface of the box.
\[ \text{Electric Flux} (kN \cdot m^2/C) \]

(b) Calculate the electric flux through the slanted surface of the box.
\[ \text{Electric Flux} (kN \cdot m^2/C) \]

(c) Calculate the electric flux through the entire surface of the box.
\[ \text{Electric Flux} (kN \cdot m^2/C) \]

Ensure to use the principles of electric flux calculation, considering the orientation of each surface with respect to the electric field direction.

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

SEE SOLUTION Check out a sample Q&A here

Step 1

VIEW

Step 2

VIEW

Step by step

Solved in 2 steps with 1 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.

Similar questions

A thundercloud produces a vertical electric field of magnitude 28.0 kN/C at ground level. You hold a 22.0 cm x 28.0 cm sheet of paper horizontally below the cloud. a. What is the electric flux through the sheet? b. What would be the flux be if you tilt the sheet of paper by 30°? c. What would be the flux be if you hold the sheet of paper vertically?
A uniform electric field of magnitude 5.7 x 104 N/C is at an angle of 10° to a square sheet with sides 5.5 m long. What is the electric flux through the sheet? Hint Electric flux is E N.m²/C.
A electric field at 3.8 cm from the centre of a long copper rod of radius 0.7 cm has a magnitude 3.6 N/C and is directed outward from the axis of the rod. Part 1 What is the charge per unit length of the copper rod? = number (rtol=0.03, atol=0) C/m Part 2 What would be the electric flux through a cube of side 8.0 cm situated such that the rod passes through the opposite sides of the cube perpendicularly? number (rtol=0.03, atol=0) Nm?/C
Consider a closed triangular box resting within a horizontal electric field of magnitude E = 7.76 x 104 N/C as shown in the figure below. 10.0 cm 30.0 cm 60.0⁰ (a) Calculate the electric flux through the vertical rectangular surface of the box. kNm²/C (b) Calculate the electric flux through the slanted surface of the box. kNm²/C Need Help? (c) Calculate the electric flux through the entire surface of the box. kN - m²/c Read It Watch It
A closed surface is in the form of a block with length c, width b and height a. In this room there is an electric field parallel to the x axis with magnitude: E = A + Bx, where A and B are constants. a. Determine the total flux over the closed surface. b. Determine the amount of charge covered by the payload. N x = a ko-
Consider a closed triangular box resting within a horizontal electric field of magnitude E = 7.86 x 104 N/C as shown in the figure below. 10.0 cm 30.0 cm 60.0⁰ E (a) Calculate the electric flux through the vertical rectangular surface of the box. kN - m²/c (b) Calculate the electric flux through the slanted surface of the box. kNm²/C (c) Calculate the electric flux through the entire surface of the box. kN - m²/c
Consider a closed triangular box resting within a horizontal electric field of magnitude E = 8.26 104 N/C as shown in the figure below. (a) Calculate the electric flux through the vertical rectangular surface of the box.