Consider a cube with edge length L immersed in a uniform electric field E along the x-direction as shown in the figure below. Calculate theelectric flux through the closed surface.yELLWe need to solve for the electric flux through each face of the cube and take the sum to calculate the net electric flux:Фnet Ф1 + Ф, +Ф; + Ф.ху+ 0xz+@yzThe flux through surface 1 is: 01=0.The flux through surface 2 is: 02=The flux through surface 3 is: 03 =The flux through the surface on the xy plane is: Oxy=The flux through the surface on the xz plane is: Oxz=The flux through the surface on the yz plane is: OyzTaking the algebraic sum, the net electric flux through the cube is: Onet =1111

Given data: The edge of the cube is, L The uniform electric field is in the positive direction of…

Consider a cube with edge length L immersed in a uniform electric field E along the x-direction as shown in the figure below. Calculate the electric flux through the closed surface. y L L We need to solve for the electric flux through each face of the cube and take the sum to calculate the net electric flux: Фрet 3 Ф1 + Ф2 + Фз + Фху + Фxz + Фуz The flux through surface 1 is: 01=0. The flux through surface 2 is: 02 = The flux through surface 3 is: O3 = The flux through the surface on the xy plane is: O The flux through the surface on the xz plane is: 0 xz= The flux through the surface on the yz plane is: Oyz= Taking the algebraic sum, the net electric flux through the cube is: Onet %3D 111

Consider a cube with edge length L immersed in a uniform electric field E along the x-direction as shown in the figure below. Calculate the electric flux through the closed surface. y L L We need to solve for the electric flux through each face of the cube and take the sum to calculate the net electric flux: Фрet 3 Ф1 + Ф2 + Фз + Фху + Фxz + Фуz The flux through surface 1 is: 01=0. The flux through surface 2 is: 02 = The flux through surface 3 is: O3 = The flux through the surface on the xy plane is: O The flux through the surface on the xz plane is: 0 xz= The flux through the surface on the yz plane is: Oyz= Taking the algebraic sum, the net electric flux through the cube is: Onet %3D 111

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

Similar questions

dx P. ++ + + + + + dq L Evaluate an expression for the electric field at point P of a uniformly charged rod with linear charge density, 2 [C/m²], as shown above. Hint: The integral will be of the form: E (P) = [ k dq , where dq = 1 dx x2 1
A uniformly charged disk of radius R=25 cm carries a total charge of Q=2 μC. R P 1 Find the magnitude of the electric field E created by the disk at the point P located at distance 40 cm from the center of the disk. E = [N/C] 2. What is the direction of the electric field E created by the disk at point P. Direction: 3. Write the expression of the electric filed E created by the disk at point P. Ē =< [N/C] 4. Find the magnitude of the electric field E created by the disk at the point P located at distance 8 m from the
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.
Consider a region of the shape of a rectangular box in the three dimensional coordinate system shown in the figure. Assuming that if there is an electric field of the form E= 3yj ,which of the following is the net electric flux through the region? 3b Заb Зас 3abc O O O OO
Find the flux in N.m².C-1 of a constant electric field E = (5.3180x10^3) i + (-5.514x10^3)j + (-1.5880x10^3) k N/C, passing through an area defined by the area vector A = (-4.7990x10^0) i + (-7.69x10^0) j + (4.2900x10^-1) k m².
Find the electric flux (in units of N.m2/C) through the surface of the rectangle ( 30 cm x 40 cm) in the figure given that E = 399.7 N/C. Normal E 0-30
Imagine that in a region in space you detect a spherical symmetric electric field that increases quadratically with the distance from the center, E (r) = a · r2 . er, where a is a constant. How is the the charge distributed in that region, i.e. find p (r). increases linearly radially outward decreases linearly radially outward uniform
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
A closed surface with dimensions a = b = 0.461 m and c = 0.7837 m is located as in the figure. The electric field throughout the region is nonuniform and defined by E (a + Bx²) î where x is in meters, a = 3 N/C, and 3 = 5 N/(C. m²). Y E C. b X What is the magnitude of the net charge enclosed by the surface? Answer in units of C. Answer in units of
Sections AB and CD of a thin non-conducting ring of radius R areuniformly (with constant linear density) charged with charge + q and−q, respectively. The points ABCD form the vertices of the square. Findthe electric field in the center of the ring.
In the figure, a point charge 2.5 C is a distance d/2 directly above the center of a square of side 4.0 mm. What is the magnitude of the electric flux through the square?
A point charge is located at the origin. Centered along the x axis is a cylindrical closed surface of radius 10 cm with one end surface located at x = 2 m and the other end surface located at x = 2.5 m. If the magnitude of the electric flux through the surface at x = 2 m is 4 N . m2 /C, what is the magnitude of the electric flux through the surface at x = 2.5 m? Select one: a. 1.8 N . m2 /C b. 2.56 N . m2 /C c. 1.0 N . m2 /C d. 4.0 N . m2 /C e. 5.0 N . m2 /C