Problem 5: A spherical, insulating shell with uniform charge density p(r) = -p is centered at the origin and has an inner radius of a and an outer radius of b. At the center of the shell is a point charge q. What is the formala for the electric field in the following three regions (a) 0
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- A point charge q=5.01012 C is placed at the center of a spherical conducting shell of inner radius 3.5 cm and outer radius 4.0 cm. The electric field just above the surface of the conductor is directed radially outward and has magnitude 8.0 N/C. (a) What is the charge density on the inner surface of the shell? (b) What is the charge density on the outer surface of the shell? (c) What is the net charge on the conductor?Charge density p Figure 2: (True,False) (a) The electric field inside the sphere is given by Ē = (7 – b) An insulating sphere of radius R has a spherical hole of radius a located within its volume and centered a distance b from the center of the sphere where a < b < R(a cross section of the sphere is shown in figure 2). The solid part of the sphere has a uniform volume charge density p: (b) The electric field inside the hole is constant and is given by : (F – 6) = (True, False) -デ+ E = Ësphere + E(-p) 3e0 3e0 3€0 - 2rb cos 0: (c) The electric field inside the sphere of radius R but outside the hole of radius a is not constant and is given by |7–6| = /r2 + b² -pa3 デー5 デ+ (True, False) - 6|3 E = Ësphere + Ē(-P) 3e0 3e0 F –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 3B p3/2 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 ÞĒ- 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 p = dV Based on the given problem, we can also say that dqenc p= - 3B 1312 dV Let us first…
- The electric field at a distance of 0.154 m from the surface of an insulating solid sphere of radius 0.385 m is 1970 N/C. Assuming that the charge on the sphere is uniformly distributed, a) what is the charge density within it? b) Calculate the electric field inside the sphere at a distance of 0.150 m from the center.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= 3/2 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 fE - 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 p= AP dV Based on the given problem, we can also say that dg enc p= dv 38 312 Let us first…Problem 2. Concentric Insulating Spheres. An insulating solid sphere, of radius a, with a uniform volume charge density p > 0, is placed concentric with an insulating spherical shell, of inner radius a and outer radius b, with a uniform volume charge density -p. Calculate the electric field at a distance r from the common center of the two spheres for (a) 0 b. b
- What is the magnitude of electric field outside and inside of a solid sphere which of radius R and has charge density p(r) = po(1 – ) for r R (outside). The total charge is Q on the sphere. for r > R; E = por( – 2AR) 4Tegr² O a. E = for r R; E = por( - 2AR) for r R; E = por² ( - ziR) for r R; E = po7²( - 2iR) for r R; E = Por( - i) for ra) A solid sphere, made of an insulating material, has a volume charge density of ? = a/r, where r is the radius from the center of the sphere, a is constant, and a > 0. What is the electric field within the sphere as a function of the radius r? Note: The volume element dV for a spherical shell of radius r and thickness dr is equal to 4?r2dr. (Use the following as necessary: a, r, and ?0.) b.) What If? What if the charge density as a function of r within the charged solid sphere is given by ? = a/r^2? Find the new magnitude and direction of the electric field within the sphere at radius r. (Use the following as necessary: a, r, and ?0.)Problem 4. Exponentially Decaying Charge Density. A spherically symmetric charge distribution is described by a volume charge density given byp = Poe¯"/a, where po and a are both positive constants, and r is the distance from the center of the distribution. (a) Determine the electric field at any value of r. (b) Sketch the graph of the electric field magnitude with r. (c) Describe how the electric field varies with the distance r.lelt Charge of uniform density (40.0 uC/m²) is distributed on a spherical surface (radius a = 3.0 cm), and a second concentric spherical surface (radius b 5.0 cm) carries a uniform charge density of (60.0 µC/m2). What is the magnitude of the electric field at a point r 4.0 cm from the center of the two surfaces? Hint: volume of a sphere =( 4Tor³)/3; area of the surface of a sphere = 4tr?. a) 11.3 x 105 N/C, b) 2.3 x 108 N/C, c) 5.0 x 105 N/C, d) 2.5 x106 N/C. e) 1.3 x 107 N/C. TOSHIBA A of & 5 6 7 7 V 8A 990 R Ty YIU قProblem 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= 1312 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 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 da p= dv Based on the given problem, we can also say that dgenc p= 3B 3/2 dv Let us first solve for…An infinițe line charge has constant charge-per-unit-length 2. Surrounding the line charge is a cylindrical shell of radius R, and carrying a constant charge-per-unit-area o. Given 2, what must o be in order to get zero electric field for all points outside the cylindrical shell? For that o, what is the electric field in between the line of charge and the shell? a = ? RSEE MORE QUESTIONS
