If the electric field at the shell’s inner wall is, 2. Suppose that the radius of the central wire is 25 mm, the inner radius of the shell 1.4 cm, and the length of the shell 16 cm. The resulting “avalanche” of electrons is collected by the wire, generating a signal that is used to record the passage of the original particle of radiation. Below examples mostly considered an electric field as a vector field. Their vector field referred here could either be a magnetic field, gravitational field or electric field. More free electrons are thereby created, and the process is repeated until the electrons reach the wire. An arbitrarily closed surface in three-dimensional space through which the flux of vector fields is determined is referred to as the Gaussian surface. However, the electric field is so intense that, between collisions with gas atoms, the free electrons gain energy sufficient to ionize these atoms also. The resulting free electrons (e) are drawn to the positive wire. A particle of radiation entering the device through the shell wall ionizes a few of the gas atoms. For Gauss’ Law we will use a closed surface like a sphere or a box or a cylinder. We can think of electric flux as the number of electric field lines passing through some surface area. The shell contains a low-pressure inert gas. The Gauss’ Law method of determining the electric field depends on the idea of electric flux. A thin, positively charged central wire is surrounded by a concentric, circular, conducting cylindrical shell with an equal negative charge, creating a strong radial electric field. Let us consider a charged rod whose flux has to be determined, so we draw a Gaussian surface or Gaussian cylinder of length l and radius r around the rod as shown below.Figure 23-61 shows a Geiger counter, a device used to detect ionizing radiation, which causes ionization of atoms. Now the D option says that the surface is closed hence it satisfies the law and it says that the electric field has a normal component which is either zero or fixed value that implies that the electric field is well defined. Hence it is not necessary that the surface will be spherical. To determine the electric field for the long wire of uniform charge we consider the cylindrical surface. \[\phi =\mathop\limits_SĪs the law states that it calculates the electric field for closed surfaces, hence option B is incorrect.įor any closed surface around the charge distribution the charge distribution may be discrete and then the electric field will not be well defined. The Gauss law states that the flux generated by an electric field in a closed surface can be given as the surface integral of the electric field, its mathematical representation is So we will discuss Gauss law and the properties of the Gaussian surface. flux passing through the surface of the gaussian sphere as a function of r for r. Here we have to define the Gaussian surface for the electric field due to a charge distribution. Calculate the electric flux through (a) the vertical rectangular surface. The Gaussian surface is associated with Gauss law. The Gaussian surface is a closed surface through which the flux is calculated for the electric or magnetic field. Now, using the concept of the net charge to be zero for a Gaussian surface, we can get the electric flux through the curved surface. By putting the given values, we can find the answer. Obtain the expression for angular velocity by differentiating the angular displacement. From the given quantities, find the time for the angular displacement. Obtain the expression for the angular displacement of the particle. Hint: Define angular simple harmonic motion.
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