**Understanding Electric Flux Through a Circular Surface**An electric field has a uniform value (doesn't change with position) that can be described by the following equation:\[ \vec{E} = \left( A \hat{j} + B \hat{k} \right) \, \text{N/C} \]where A and B are given by the values below.- \( A = 1.35 \)- \( B = 1.80 \)There is a flat circular surface that is in the x-y plane and centered at the origin point (0,0,0). This surface has a radius of \( 1.65 \) m.**Task:**Calculate the magnitude of the electric flux through the surface due to the electric field described above. Your units should be \( \text{Nm}^2/\text{C} \).**Notes:**- The electric field vector \( \vec{E} \) is composed of the y-component (\( \hat{j} \)) with magnitude \( A = 1.35 \, \text{N/C} \) and the z-component (\( \hat{k} \)) with magnitude \( B = 1.80 \, \text{N/C} \).- Since the surface is in the x-y plane, the vector normal to this surface (\( \vec{A} \)) points in the z-direction.To understand the electric flux (\( \Phi_E \)) calculation, consider the following step:1. The electric flux through a surface is given by the dot product of the electric field vector \( \vec{E} \) and the area vector \( \vec{A} \): \[\Phi_E = \vec{E} \cdot \vec{A}\]2. For a circular surface with radius \( r = 1.65 \) m in the x-y plane, the area vector \( \vec{A} \) has a magnitude equal to the area of the circle and points in the positive z-direction (\( \hat{k} \)): \[ A = \pi r^2 = \pi (1.65)^2 \approx 8.55 \, \text{m}^2 \] Therefore, \( \vec{A} = 8.55 \hat{k} \, \text{m}^2

Given the value of electric field is given by E = (A j + B k) N/C where A = 1.35 and B= 1.80 Radius…

An electric field has a uniform value (doesn't change with position) that can be described by the following equation: E = (Aĵ + B k) N/C where A and B are given by the values below. A = 1.35 B = 1.80 There is a flat circular surface that is in the x-y plane and centered at the origin point (0,0,0). This surface has a radius of 1.65 m. Calculate the magnitude of the electric flux through the surface due to the electric field described above. Your units should be Nm²/C.

An electric field has a uniform value (doesn't change with position) that can be described by the following equation: E = (Aĵ + B k) N/C where A and B are given by the values below. A = 1.35 B = 1.80 There is a flat circular surface that is in the x-y plane and centered at the origin point (0,0,0). This surface has a radius of 1.65 m. Calculate the magnitude of the electric flux through the surface due to the electric field described above. Your units should be Nm²/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: am stuck on how to solve this problem. Find the electric field at the location of qa in the figure…

Q: Part (a) What is the change in electric potential, in volts, from point (0, 0) to point (x₁, 0)? V…

Q: ) An electric charge q is placed at the origin in 3-space. The resulting electric r field E is given…

Q: A nonuniform electric field is given by the expression E = ay i+ bzj + cx k, where a, b, and care…

A: Electric field is Here are constants.Rectangular surface in xy-plane, whose dimensions are from to…

Q: Three very lage sre planes of charge nged houn on ede he fgure oFrom le rg he piares have charge…

A: σ1 = -0.51 μCm2σ2 = 0.27 μCm2σ3 = -0.39 μCm2

Q: When E and A were parallel, we called the quantity EA the electric flux through the surface. For the…

A: Required to rewrite the quantity described as a product of E and A.

Q: Ē= This problem checks your understanding of the term in the equation for the electric field due to…

A: Given data: A charged particle at a point S coordinates is (1m, 6m, 10m). The electric field vector…

Q: In a given region of empty space, By = B0 * e-t / T (where B0 and T are constants), but Bx and Bz…

A: X component of electric field is time and position dependent.Units on both sides make…

Q: Problem 3: Imagine you have a very thin plastic hoop of diameter D, which is charged up to total net…

A: A point's electric field occurs when a charge of any kind is present in that point's space. electric…

Q: Part E What is the electric flux through the cube face S3? Express your answer in newton times…

Q: An electric field is 62 V/m at 22 degrees down from the due west with a displacement ( delta r ) of…

Q: Use Gauss’ Law to write an equation for the electric field at a distance R1 < r < R2 from the center…

Q: A sphere of radius R has total charge Q. The volume charge density (C/m³) within the sphere is p(r)…

Q: A uniform electric field of magnitude E = 460 N/C makes an angle of ? = 65.5° with a plane surface…

Q: Problem 2: A closed hollow cylinder (i.e., with capped ends) is situated in an electric field…

Q: Figure 1 of 1 What is the electric field at point 1 in (Figure 1)? Assume that a = 2.0 cm, b = 1.0…

Q: What would be the electric field for z0 > R for the following integral? The integral is the answer…

Q: A constant electric field 5.8 N/C is pointing in the positive x-direction. Calculate the magnitude…

Q: What is the electric field at the center of a sphere of radius R whose volume is uniformly filled…

A: Given charge Q is uniformly distributed over the volume of sphere having radius R. Therefore charge…

Q: this question, Figure 2 (see image). Consider the electric field of a disk of radius R and surface…

Q: A closed surface with dimensions a = b = 0.400 m and c = 0.800 m is located as shown in the figure…

Q: An electric field E = x (-1.12 × 10³ N · m²/C) is 25.0 cm due to this charge distribution. UC is…

Q: Two parallel conducting plates, each of cross sectional area 540 cm2, are 8.0 cm apart and…

A: Given: cross section area A = 540cm2=540*10-4m2 number of electrons n=5.6×1012 (a) total charge Q on…

Q: Find the flux of F=−zi+yj−xk out of a sphere of radius 4 centered at the origin. Hint: Your answer…

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

**Understanding Electric Flux Through a Circular Surface**

An electric field has a uniform value (doesn't change with position) that can be described by the following equation:

\[ \vec{E} = \left( A \hat{j} + B \hat{k} \right) \, \text{N/C} \]

where A and B are given by the values below.

- \( A = 1.35 \)
- \( B = 1.80 \)

There is a flat circular surface that is in the x-y plane and centered at the origin point (0,0,0). This surface has a radius of \( 1.65 \) m.

**Task:**
Calculate the magnitude of the electric flux through the surface due to the electric field described above. Your units should be \( \text{Nm}^2/\text{C} \).

**Notes:**
- The electric field vector \( \vec{E} \) is composed of the y-component (\( \hat{j} \)) with magnitude \( A = 1.35 \, \text{N/C} \) and the z-component (\( \hat{k} \)) with magnitude \( B = 1.80 \, \text{N/C} \).
- Since the surface is in the x-y plane, the vector normal to this surface (\( \vec{A} \)) points in the z-direction.

To understand the electric flux (\( \Phi_E \)) calculation, consider the following step:

1. The electric flux through a surface is given by the dot product of the electric field vector \( \vec{E} \) and the area vector \( \vec{A} \):

\[\Phi_E = \vec{E} \cdot \vec{A}\]

2. For a circular surface with radius \( r = 1.65 \) m in the x-y plane, the area vector \( \vec{A} \) has a magnitude equal to the area of the circle and points in the positive z-direction (\( \hat{k} \)):

\[
A = \pi r^2 = \pi (1.65)^2 \approx 8.55 \, \text{m}^2
\]

Therefore, \( \vec{A} = 8.55 \hat{k} \, \text{m}^2

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!

SEE SOLUTION Check out a sample Q&A here

Step 1

VIEW

Step 2

VIEW

Trending now

This is a popular solution!

Step by step

Solved in 2 steps

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

Question 5: Consider the following situation: A ring of radius R is centered on the x axis a distance l from the origin. The ring has a uniform electric charge density per unit length d, d > 0. Show that the electric filed at a location on the x axis, a distance l from the origin comes out to be Ē = 2nkeAlR (?+ R?}3/2 Hint: You will need to set up an integration for this problem.
Question 1 Four stationary electric charges produce an electric field in space. The electric field depends on the magnitude of the test charge used to trace the field O has different magnitudes but same direction everywhere in space is constant everywhere in space has different magnitude and different directions everywhere in space CANAD
E ↑ Απεργια This problem checks your understanding of the term in the equation for the electric field due to a point charge, Consider a charged particle at a point S whose coordinates are (4 m, 2 m, 6 m). We would like to find the electric field vector at a point P whose coordinates are (7 m, 8 m, 1 m). The "unit vector" is a vector that points from S to P that has length of 1 (or "unity"). What is its y component, in meters? (Report your answer to one decimal place.)
In the figure below, q1 = 1.50××10^−7 C and q2 = 3.50×10^−7 C. 1)What is the x component of the electric field E⃗ E→ at the point (x, y) = (6.00 m, 3.00 m)? (Express your answer to three significant figures in N/C.) NOTE -- the figure is correctlylabeled in centimeters and the question is asking for the E-field at (6,3) meters. Account for this in your calculation. 2)What is the y component of the electric field E⃗ E→ at the point (x, y) = (6.00 m, 3.00 m)? (Express your answer to three significant figures in N/C.)
A cube has sides of length L = 0.370 m. It is placed with one corner at the origin as shown in the figure (Figure 1). The electric field is not uniform but is given by E =(-5.69 N/(C. m) )æi+(3.20 N/(C - m) )zk. Part A Find the electric flux through each of the six cube faces S1, S2, S3, S4, S5. and S6. Enter your answers numerically separated by commas. Eν ΑΣφ ? d1, Þ2, Þ3 , 4 , Þ5 , Þ6 = (N/C) - m² Submit Previous Answers Request Answer Part B Find the total electric charge inside the cube. ΑΣφ ? q = Figure 1 of 1 Submit Request Answer S2 (top) So (back) Provide Feedback S3 (right side) (left side) L L L. S, (bottom) S, (front)
We have calculated the electric field due to a uniformly charged disk of radius R, along its axis. Note that the final result does not contain the integration variable r: R. Q/A 2€0 Edisk (x² +R*)* Edisk perpendicular to the center of the disk Uniform Q over area A (A=RR²) Show that at a perpendicular distance R from the center of a uniformly negatively charged disk of CA and is directed toward the disk: Q/A radius R, the electric field is 0.3- 2€0 4.4.1b
A cube has sides of length L = 0.370 m. It is placed with one corner at the origin as shown in the figure (Figure 1). The electric field is not uniform but is given by E =( -5.69 N/(C. m) )æi+( 3.20 N/(C · m) )zk. Part A Find the electric flux through each of the six cube faces Sı, S2, S3, S4, S5, and Sg- Enter your answers numerically separated by commas. ν ΑΣφ D1, Þ2 , D3 , Þ4 , Þ5 , Þ6 = (N/C) - m² Submit Request Answer Part B Find the total electric charge inside the cube. q = Figure 1 of 1 Submit Request Answer S2 (top) S6 (back) Provide Feedback S1 (left side) L - S3 (right side) y L L. S4 (bottom) S3 (front)
There are three static charges (Q1 = 4 nC, Q2 = - 4 nCn and Q3 = - 4nC) located in different locations of Fig. 1. Calculate the net electric field at the given point P.
A non-conducting rod of length with a uniform charge density λ and a total charge is lying along the x-axis, as illustrated in Figure. point P, located at a distance y off the axis of the rod. Compute the electric field at a y P 8: x
An insulating sphere of radius R = 3 cm has positive charge uniformly distributed throughout its entire volume. The electric field at the surface of the sphere has a magnitude of E = 5x107 V/m (a) [3 points] Calculate the volumetric charge density p of the sphere (in C/m3) and Calculate the magnitude of the electric field at a point located atr= 1cm, inside the insulator.
Needs Complete typed solution with 100 % accuracy.