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Find the electric field a distance *z *from the center of a spherical surfaceof radius *R *(Fig. 2.11) that carries a uniform charge density
*chargeq *on the sphere. [*Hint: *Use the law of cosines to write in terms of R and

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# Chapter 2 Solutions

Introduction to Electrodynamics

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- 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 using superposition.
*arrow_forward*Do asap*arrow_forward*Problem 2.7 Find the electric field a distance z from the center of a spherical surface of radius R (Fig. 2.11), which carries a uniform charge density o. Treat the case z R (outside). Express your answers in terms of the total charge q on the sphere. [Hint: Use the law of cosines to write r in terms of R and 0. Be sure to take the positive square root: VR? + z? - 2Rz = (R – z) if R > z, but it's (z – R) if R < z.]*arrow_forward* - Just 6.21 a and b
*arrow_forward*A solid sphere of radius a, centered at the origin, carries a volume charge density of p(r) = ar², where a is a constant (see Fig. 2 below). a. Calculate the electric field inside and outside of the sphere. (NOTE: For a spherically symmetric system, dV = r² sin(e) de dødr. Therefore, the volume integral of some function p(r) is given by fp(r)r² sin(e) dedodr = 4n fp(r)r²dr) b. Plot the resulting electric field as a function of distance from the origin. a*arrow_forward*2.6.1 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 this*arrow_forward* - Figure 1.52 shows a spherical shell of charge, of radiusa 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 using direct integration.
*arrow_forward*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 p= 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 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 AP Based on the given problem, we can also say that dgenc 3B p= dv 3/2*arrow_forward*A cylindrically symmetric charge distribution has charge density p(r) given by P()= 2na erla where Op and a are constants. Calculate the total charge O contained within a cylinder of length I = 5a and radius 2a, centred on the z-axis, and grve your answer by entering numbers in both of the boxes in the equation below: [Note that you must enter a value in each of the boxes, including 1 or 0 if appropriate. Blank boxes will be marked as incorrect.]*arrow_forward*

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