If the electric field of a point charge were proportional to 1/r3instead of 1/r2, would Gauss’s law still be valid? Explain your reasoning.(Hint: Consider a spherical Gaussian surface centered on a singlepoint charge.)

Consider Gauss's law: f Ē - d.Ã=q/€0 . Which of the following is true? E must be the electric field due to the enclosed charge If q = 0, then E = 0 everywhere on the Gaussian surface If the three particles inside have charges of +q, +q, and -2q, then the integral is zero If a charge is placed outside the surface, then it cannot affect E at any point on the surface on the surface E is everywhere parallel to dA

Gauss’ Law has an important practical application. Imagine a solid conducting object which may be in a strong electric field. If all excess charge is on the surface of the conductor and if there is no electric field inside the conductor, then it should be possible to remove any conducting material from the interior without changing the location of the charges or the electric field inside. This will leave a hollow metallic shell which has no electric field inside it. There is no electric field inside even if there is a strong field outside: the conducting shell acts as a “shield” against electric fields. Furthermore, the hollow shell can have holes in it or even be made of a wire mesh and still retain its shielding property. Such a space surrounded by a conductor is called a “Faraday cage”. Materials : • A metal cup • A hard plastic rod and fur • A metallic ball suspended by a thread Now try the experiment without touching the can with your hand or any other electrical conductor.…

The figure below shows a closed cylinder with cross-sectional area A = 2.00m2. The constant electric field E has magnitude 3.50 X 103 N/C and is directed vertically upward, perpendicular to the cylinder's top and bottom surfaces so that no field lines pass through the curved surface. Calculate the electric flux through the cylinder's (a) top and (b) bottom surfaces. (c) Determine the amount of charge inside the cylinder

Gauss’ Law has an important practical application. Imagine a solid conducting object which may be in a strong electric field. If all excess charge is on the surface of the conductor and if there is no electric field inside the conductor, then it should be possible to remove any conducting material from the interior without changing the location of the charges or the electric field inside. This will leave a hollow metallic shell which has no electric field inside it. There is no electric field inside even if there is a strong field outside: the conducting shell acts as a “shield” against electric fields. Furthermore, the hollow shell can have holes in it or even be made of a wire mesh and still retain its shielding property. Such a space surrounded by a conductor is called a “Faraday cage”. Materials : • A metal cup • A hard plastic rod and fur • A metallic ball suspended by a thread A. Hold the rod in its position outside the cup as near as possible to the ball inside the cup.…

A solid, insulating sphere of radius a has a uniform charge density throughout its volume and a total charge Q. Concentric with this sphere is an uncharged, conducting, hollow sphere whose inner and outer radii are b and c as shown. We wish to understand completely the charges and electric fields at all locations. (a) Find the charge contained within a sphere of radius r < a. (b) From this value, find the magnitude of the electric field for r < a. (c) What charge is contained within a sphere of radius r when a < r < b? (d) From this value, find the magnitude of the electric field for r when a < r < b. (e) Now consider r when b < r < c. What is the magnitude of the electric field for this range of values of r ? (f) From this value, what must be the charge on the inner surface of the hollow sphere? (g) From part (f), what must be the charge on the outer surface of the hollow sphere? (h) Consider the three spherical surfaces of radii a, b, and c. Which of these…

(a) Define uniform and non-uniform electric flux passing through a closed surface. (a) Explain Gauss's Law for Point Charges Inside Spherical and Nonspherical Surfaces, followed by the general form of Gauss's law.

The electric field everywhere on the surface of a thin spherical shell of radius 0.750 m is measured to be equal to 890 N/C and point radially toward the center of the sphere. (a) What is the net charge within the sphere's surface? (b) What can you conclude about the nature and distribution of the charge inside the spherical shell?

Consider two parallel copper plates both 1m? in area. The lower plate is charged with 4µC of charge and the upper plate with -4µC of charge. (a) Determine the charge density o on each plate. (b) Neglecting edge effects (assuming a homogeneous field between the plates), show that the electric field can be expressed in the form E = Eoê using Gauss' Law. Determine the value of Eo. (c) Consider now the addition of an open hemispherical surface (no base) of radius R between the two plates in the orientation shown in Fig. 1. What is the expression for electric flux through this surface? Given R = 3cm, what is the value of the flux? R FIGURE 1 (d) Consider now a rotation of the hemispherical surface of 90° about the x-axis. What is the the new flux through the surface? (e) Following (d), a point charge q is placed at the origin of the hemispherical surface. Assume that small enough that it does not affect the charge density on the plates. What is the new flux through the hemisphere? is