![]() The net flux of a uniform electric field through a closed surface is zero. To quantify this idea, Figure 6.4(a) shows a planar surface A+0+0+0+0=0. Again, flux is a general concept we can also use it to describe the amount of sunlight hitting a solar panel or the amount of energy a telescope receives from a distant star, for example. Similarly, the amount of flow through the hoop depends on the strength of the current and the size of the hoop. As you change the angle of the hoop relative to the direction of the current, more or less of the flow will go through the hoop. This tutorial aims to provide the most concise possible insight on finding electric flux in three different situations while still providing the core necessary ideas. The numerical value of the electric flux depends on the magnitudes of the electric field and the area, as well as the relative orientation of the area with respect to the direction of the electric field.Ī macroscopic analogy that might help you imagine this is to put a hula hoop in a flowing river. Having to find the electric flux through an open or closed surface can pose a huge challenge for physics students. A cube whose sides are of length d is placed in a uniform electric field of magnitude \(\displaystyle E=4.Figure 6.3 The flux of an electric field through the shaded area captures information about the “number” of electric field lines passing through the area. What is the net charge enclosed by the surface?ģ8. The electric flux through a spherical surface is \(\displaystyle 4.0×10^4N⋅m^2/C\). ![]() This is because, any field line from the. What is the total charge enclosed by the box?ģ7. When a charge is moved outside a closed surface, the net (outward) flux passing through the surface becomes zero. The electric flux through a cubical box 8.0 cm on a side is \(\displaystyle 1.2×10^3N⋅m^2/C\). Find the magnitude of the electric flux through the shaded face due to q. A charge q is placed at one of the corners of a cube of side a, as shown below. ![]() (b) How precisely can we determine the location of the charge from this information?ģ5. (a) How much charge is inside the sphere? A net flux of \(\displaystyle 1.0×10^4N⋅^m2/C\) passes inward through the surface of a sphere of radius 5 cm. Find the net electric flux though the surfaces of the cube.ģ4. A point charge of \(\displaystyle 10μC\) is at an unspecified location inside a cube of side 2 cm. Mathematically, electrical flux is the product of the electric field (E), surface area (A) and the cosine of the angle. If there are no other charges in this system, what is the electric flux through one face of the cube?ģ3. A point charge q is located at the center of a cube whose sides are of length a. Find the electric flux through the closed surface whose cross-sections are shown below.ģ2. ![]() In pictorial form, this electric field is shown as a dot, the charge, radiating lines of flux. Determine the electric flux through each closed surface whose cross-section inside the surface is shown below.ģ1. Gauss’s law states that the net electric flux through any hypothetical closed surface is equal to 1/0 times the net electric charge within that closed surface. What is the flux through the surface due to the electric field of the charged wire?ģ0. An infinite charged wire with charge per unit length \(\displaystyle λ\) lies along the central axis of a cylindrical surface of radius r and length l. Repeat the previous problem, given that the circular area is (a) in the yz-plane and (b) 45° above the xy-plane.Ģ9. What is its electric flux through a circular area of radius 2.0 m that lies in the xy-plane?Ģ8. ![]()
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |