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Here’s an alternative derivation of Eq. 3.10 (the surface charge density induced on a grounded
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- Problem An insulating sphere with radius R contains a total non-uniform charge (i.e. Hydrogen atom) Q such that its volume charge density is 38 312 where Bis a constant and ris the distance from the center of the sphere. What is electric field at any point inside the sphere? Solution To find the electric field inside the non-uniformly charged sphere, we may apply integration method or the Gauss's Law method. Here, let us use the Gauss's law which is expressed as We will choose a symmetric Gaussian surface, which is the surface of a sphere, then evaluate the dot product to obcain (Equation 1) The issue however is how much charge does the Gaussian surface encoses? Since, our sphere is an insulating material. charges will get distributed non-uniformly within the volume of the object. So, we look into the definition of volume charge density to find the enclosed charge. So, we have dq p= dv Based on the given problem, we can also say that dgenc p= 38 dVarrow_forwardConsider linto. two The sphenical conductor (radius a) separetea latikude of ks from two parts at morth. The Parts. are insulated another One part (latitude s45') is top Potential of V and -V In. held at a bottom part Clattitude sphenical cocrdinates, take zaxis termss the at the of angle es painting where to north. In the two parts? Determine the entre are the electric potential space outside the Conductor.arrow_forward(ITV) (a) Solve the complete field solutions for the TE11 mode. At what location(s) on the x– y plane does the electric field have the maximum strength? (b) For the TE10 mode, calculate both the phase velocity v, and the group velocity vg, as well as their product v,*Vg. In HW V.2, we calculated the velocity of the energy propagation. Compare it to your answers and what can you conclude? Consider a square waveguide with all sides of equal length a.arrow_forward
- An Electron moves in the presence of a uniform magnesic field in The Z -direchin (B=Bk). Part a: write do wn a Vector potinh al A that generates the Specific magnhic field B. Assume A is independent of time and Az=0 Part b: Write down the darteoi ham iltonian for the system in terms of TTi , Where Tl; = P; -eAilc' is and e the electron Charge· * Part C:Compute [ TTe;T ly]arrow_forwardProblem An insulating sphere with radius R contains a total non-uniform charge (i.e. Hydrogen atom) Q such that its volume charge density is 3B p = 3/2 where B is a constant and r is the distance from the center of the sphere. What is electric field at any point inside the sphere? Solution To find the electric field inside the non-uniformly charged sphere, we may apply integration method or the Gauss's Law method. Here, let us use the Gauss's law which is expressed as PĒ - dÃ= We will choose a symmetric Gaussian surface, which is the surface of a sphere, then evaluate the dot product to obtain A = (Equation 1) The issue however is how much charge does the Gaussian surface encloses? Since, our sphere is an insulating material, charges will get distributed non-uniformly within the volume of the object. So, we look into the definition of volume charge density to find the enclosed charge. So, we have dq dV Based on the given problem, we can also say that dq enc p= 3B 13/2 dV Let us first…arrow_forwardFigure 1.52 shows a spherical shell of charge, of radiusa and surface density σ, from which a small circular piece of radius b << ahas been removed. What is the direction and magnitude of the fieldat the midpoint of the aperture? Solve this exercise using the relationship for a force on a small patch.arrow_forward
- Using the method of integration, what is the electric field of a uniformly charged thin circular plate (with radius R and total charge Q) at xo distance from its center? (Consider that the surface of the plate lies in the yz plane) Solution A perfect approach to this is to first obtain the E-field produced by an infinitesimal charge component of the charge Q. There will be several approaches to do this, but the most familiar to us is to obtain a very small shape that could easily represent our circular plane. That shape would be a ring. So for a ring whose charge is q, we recall that the electric field it produces at distance x0 is given by E = (1/ )(x0q)/( 242) Since, the actual ring (whose charge is dq) we will be dealing with is an infinitesimal part of the circular plane, then, its infinitesimal electric field contribution is expressed as = (1/ )(x0 2. ) We wish to obtain the complete electric field contribution from the above equation, so we integrate it from 0 to R to obtain E =…arrow_forwardC harge below. The. A neutral Conductor with a Cavity has a placed in its cavity as shown As shon electric flux the gavosian through Sur face enclosing the Conductor (as, shoon n fighre) EE=&E-JA =1 (10) N is found to be (a) Find the electric charge of t. (b) Fnd the Cmount of Charge on the inner Surface of the conductor. () Find the amount of Charge on the outkr Surface of the Conductor. Graussian Surfacearrow_forward2.6.1 Divergence Find hv.Edz, where E is the electric field due to a point charge q at x' and V is a sphere centered on the origin with radius b> |x'). please solve thisarrow_forward
- Figure 1.52 shows a spherical shell of charge, of radius a and surface density σ, from which a small circular piece of radius b << a has been removed. What is the direction and magnitude of the field at the midpoint of the aperture? Solve this exercise in three ways: a) direct integration, b) by superposition, and c) using the relationship for a force on a small patch.arrow_forwardAn electric force is expressed as F = -pz sin o a, + p cos pa,. Transform F into cartesian coordinates and compute its magnitude at point (2,-1).arrow_forwardQuestion 2 a. Electrostatic fields have many applications, but most are low “low-power" applications; that is relatively low forces are involved. Explain why this is so. b. Given that D= (10ră ) ar (c/m²) in cylindrical coordinates, evaluate both sides of the divergence theorem for the volume enclosed by r= Im, r= 2m and Z= 0 and Z= 10arrow_forward
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