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The trapezoid shape below is comprised of 6 surfaces. It lies in a region when an electric field ...
Problem A.1 - Calculate electric flux f5) The electric field due to an infinite line of charge is perpendicular to the line and has magnitude E . Consider an imaginary cylinder with radius e-25 cm and length L = 40 cm that has an infinite line of positive charge running along its axis. The charge per unit length is 3 HC/m. Do not use Gauss's Law, but actually calculate the flux! a) What is the electric flux through the cylinder...
A cube of side L-3.2 m lies in a region where the electric field is given by E-2 7+5.4)i -4.0k N/C. We wish to find the net electric flux through the cube by first calculating the flux through each of (a) What is the flux through the left face of the cube? N-m2/C (b) What is the flux through the right face of the cube? N m2/c (c) What is the flux through the top face of the cube? (d)...
3. a) Find the net electric flux through this cube. b) Explain why we can't use Gauss's Law to find the electric field here. -8.00 nC +3.00 nC c) Is there a shape to which Gauss's Law could be applied in order to solve for the electric field of this double point charge situation?
A 10.0-cm diameter hemisphere lies within a uniform electric field of magnitude 1000NC1000NC and direction perpendicular to the flat side of the hemisphere. The hemisphere contains no charge, and the flux through the base is in the outward direction. The flux through the rounded portion of the surface is
PRACTICE: Some shape is made up of 8 different surfaces. If the electric flux through each surface of the shape is given below, how much charge is enclosed by this shape? 01 = 50 Nm IC 03 = -150 Nm+1C 05 = -45 Nm2/C 07 = 100 Nm2/C = 75 Nm2/C Q= 0 Nm2/C = 150 Nm2/C Q=-125 Nm2/C
A hollow cube is immersed in a electric field parallel to the x axis. at all points of the front surface the E-field is -34i N/C and at the back surface the E-field is +20i N/C. find the net charge inside the cube. so I get that there is effectively no flux through the top and side surfaces but im not really sure how to calculate the flux through the front/back without being provided the dimensions of the cube, would...
6. The electric field in the region of space shown is given by E (8i+2yj) N/C where y is in m. What is the magnitude of the electric flux through the top face of the cube shown? 2n 7. a a Charge of uniform linear density (4.0 nC/m) is distributed along the entire x axis. Determine the magnitude of the electric field on the y axis at y-2.5 m.
Which of the following statements about Gauss's law are correct? (There may be more than one correct choice.)Choose all that apply.The electric flux passing through a Gaussian surface depends only on the amount of charge inside that surface, not on its size or shape.Gauss's law is valid only for symmetric charge distributions, such as spheres and cylinders.If there is no charge inside of a Gaussian surface, the electric field must be zero at points of that surface.Only charge enclosed within...
A 3.0 cm diameter circle lies in the xz plane in a region where the electric field vector is given by the Cartesian coordinate expression: E = (1500i+ 1500j -1500k) N/C where i, j, k are the unit vectors. Draw the situation described above. Which components of the electric field contribute to the electric flux through the circle? What is the electric flux through the circle?
The figure below represents the top view of a cubic surface in a uniform electric field oriented parallel to the top and bottom faces of the cube. The field makes an angle θ=40o with side and the area of each face is A . a) Draw an area vector A for each face, , , , b) Label the angle between the area vector A and the electric field vector E for each face. c) Find flux through each face....