Assignment Gauss's law and Applications

Michael Kurniawan 1910403011 University of Cincinnati GAUSS'S LAW AND APPLICATIONS 1. What is Gauss's law in electrostatics? Gauss's law in electrostatics states that the total electric flux through a closed surface is equal to the net charge enclosed within the surface divided by the permittivity of free space (𝜖0). Mathematically, it is expressed as: where Φ𝐸 is the electric flux, 𝐸 is the electric field, 𝑑𝐴 is the differential area vector on the closed surface, and 𝑄enc is the net charge enclosed by the surface 2. How do you write the integral form of Gauss's law? The integral form of Gauss's law is: This equation states that the surface integral of the electric field 𝐸 over a closed surface (also known as a Gaussian surface) is equal to the net charge 𝑄enc enclosed by the surface divided by the permittivity of free space 𝜖0. The dot product 𝐸⋅𝑑𝐴 represents the component of the electric field passing through the area element 𝑑𝐴 perpendicular to the surface. 3. How is Gauss's law used to calculate the electric field around a point charge? To use Gauss's law to calculate the electric field around a point charge 𝑄, you choose a spherical Gaussian surface centered on the charge with radius 𝑟. Due to symmetry, the electric field 𝐸 is radially outward and has the same magnitude at all points on the surface. The flux through the surface is : According to Gauss's law: Equating the two expressions for flux: Solving for 𝐸, the electric field at a distance 𝑟 from a point charge 𝑄 is : This is the same result obtained from Coulomb's law 4. What is electric flux and how is it calculated? Electric flux (Φ𝐸) is a measure of the number of electric field lines passing through a given surface. It is calculated as the surface integral of the electric field (𝐸) over the area (𝐴) of the surface. Mathematically, it is expressed as: where 𝐸 is the electric field vector, 𝑑𝐴 is the differential area vector perpendicular to the surface, and the dot product 𝐸⋅𝑑𝐴 accounts for the component of the electric field passing through the area element. 5. How is Gauss's law applied to spherical symmetry? Gauss's law is particularly useful in calculating the electric field for systems with spherical symmetry, such as a point charge or a uniformly charged spherical shell. For a point charge 𝑄 at the center of a spherical Gaussian surface with radius 𝑟: a. The electric field 𝐸 is radially symmetric and has the same magnitude at every point on the Gaussian surface. b. The flux through the surface is: c. According to Gauss's law: d. Equating the two expressions for flux: e. Solving for 𝐸, the electric field at a distance 𝑟 from the center is: For a uniformly charged spherical shell, the electric field inside the shell (for 𝑟<𝑅, where 𝑅 is the radius of the shell) is zero, and outside the shell (for 𝑟>𝑅) it is as if all the charge were concentrated at the center. 6. How does Gauss's law help in determining the electric field around a long, charged wire (cylindrical symmetry)? For a long, uniformly charged wire with linear charge density 𝜆, Gauss's law can be used to calculate the electric field in the region around the wire: a. Choose a cylindrical Gaussian surface with radius 𝑟 and length 𝐿 coaxial with the wire. b. Due to cylindrical symmetry, the electric field 𝐸 is radially outward and has the same magnitude at every point on the cylindrical surface. c. The flux through the cylindrical surface is: d. According to Gauss's law: where 𝑄enc=𝜆𝐿 is the charge enclosed by the Gaussian surface. e. Equating the two expressions for flux: f. Solving for 𝐸, the electric field at a distance 𝑟 from the wire is: 7. What happens to the electric field inside a hollow conductor according to Gauss's law? According to Gauss's law, the electric field inside a hollow conductor is zero when the conductor is in electrostatic equilibrium. This is because the charges on a conductor rearrange themselves on the surface in such a way that they cancel any internal electric fields. By choosing a Gaussian surface entirely inside the hollow region of the conductor, we can see that there is no net charge enclosed (𝑄enc=0), leading to: Since the integral of 𝐸⋅𝑑𝐴 is zero, the electric field 𝐸 inside the conductor must also be zero. 8. How is Gauss's law used to calculate the electric field near a flat, charged sheet (planar symmetry)? For an infinite, uniformly charged sheet with surface charge density 𝜎, Gauss's law can be applied to calculate the electric field: a. Choose a Gaussian "pillbox" that straddles the sheet, with its flat faces parallel to the sheet and perpendicular to the electric field. b. The electric field 𝐸 is perpendicular to the sheet and has the same magnitude on both sides of the sheet. c. The flux through the pillbox is: where 𝐴 is the area of the pillbox faces. d. According to Gauss's law: where 𝑄enc=𝜎𝐴 is the charge enclosed by the Gaussian surface e. Equating the two expressions for flux: f. Solving for 𝐸, the electric field near the charged sheet is: 9. What is the effect of a dielectric on Gauss's law and the calculation of the electric field? A dielectric material affects the electric field and Gauss's law by reducing the effective field within the material. When a dielectric is introduced, it becomes polarized, reducing the net electric field. Gauss's law in the presence of a dielectric is modified to include the permittivity of the dielectric material (𝜖): where 𝐷=𝜖𝐸 is the electric displacement field and 𝑄free is the free charge enclosed by the Gaussian surface. The electric field 𝐸 in the presence of a dielectric is given by: This shows that the electric field is reduced by the factor of the relative permittivity (𝜅=𝜖/𝜖0) of the dielectric. 10. How does Gauss's law explain the distribution of charge on the surface of a conductor? Gauss's law explains that in electrostatic equilibrium, the electric field inside a conductor is zero. As a result, any excess charge on a conductor must reside on its surface. If we choose a Gaussian surface just inside the conductor's surface, the enclosed charge is zero because there is no electric field inside the conductor. Consequently, any net charge must be on the outer surface. The charge distributes itself in such a way as to ensure that the electric field inside remains zero and the electric field just outside the surface is perpendicular to the surface, with magnitude: where 𝜎 is the surface charge density. This distribution minimizes the repulsive forces between like charges on the conductor's surface