Thus, B can then be expressed asB=Q/Now we are ready to solve for the charged enclosed by the Gaussian surface.We apply the same definition of volume charge density, to obtain the integral3BdV9enc'o p3/2the difference is now, the limits of r will be from 0 to r. Evaluating the integral and simplifying, we obtainlenc =By substitution to Equation 1 above, then using A =and simplifying, we obtainE = (1/(Qr/R An insulating sphere with radius R contains a total non-uniform charge (i.e. Hydrogen atom) Q such that its volume charge density is3Bp=p312where 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?SolutionTo 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 asfE - dÃ= Q/ epsilono%|We will choose a symmetric Gaussian surface, which is the surface of a sphere, then evaluate the dot product to obtainA =(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 volumecharge density to find the enclosed charge. So, we havedqp=dVBased on the given problem, we can also say thatdqenc3Bp=dV1312Let us first solve for B.Our enclosed charge would have limits from 0 to Q. Then r would have the limits 0 to R. Thus, the equation above becomesR38dVp3/2where dV is the infinitesimal volume.By evaluating the integral and simplifying, we obtain the followingRfor the limits from 0 to R%3D

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= 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?

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= 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?

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter19: Electric Forces And Electric Fields

Section: Chapter Questions

Problem 50P: An insulating solid sphere of radius a has a uniform volume charge density and carries a total...

See similar textbooks

Similar questions

A charge distribution that is spherically symmetric but not uniform radially produces an electric field of magnitude E = Kr4, directed radially outward from the center of the sphere. Here r is the radial distance from that center, and K is a constant.What is the volume density r of the charge distribution?
Consider two thin disks, of negligible thickness, of radius R oriented perpendicular to the x axis such that the x axis runs through the center of each disk. The disk centered at x=0 has positive charge density η, and the disk centered at x=a has negative charge density −η, where the charge density is charge per unit area. What is the magnitude E of the electric field at the point on the x axis with x coordinate a/2? Express your answer in terms of η, R, a, and the permittivity of free space ϵ0.
A solid insulating sphere of radius 0.07 m carries a total charge of 25 uC. Concentric with this sphere is a conducting spherical shell of inner radius 0.12 m and outer radius of 0.18 m, and carrying total charge of -54 uC. Find (a) the charge distribution for the insulating sphere and the conducting spherical shell, and the magnitude of the electric field at the follwing distances from the center of the two spheres and shell: (b) 0.05 m, (c) 0.10 m, (d) 0.15 m, and (e) 0.25 m.
Figure 2 shows a nonconducting rod with a uniformly distributed charge Q. The rod forms a half-circle with radius R and produces an electric field of magnitude Earc at its center of curvature P. If the arc is collapsed to a point at distance R from P, by what factor is the magnitude of the electric field at P multiplied?
A solid nonconducting sphere of radius R = 6.7 cm has a nonuniform charge distribution of volume charge density ρ = (16.7 pC/m3)r/R, where r is radial distance from the sphere's center. (a) What is the sphere's total charge? What is the magnitude E of the electric field at (b) r = 0, (c) r = R/2.0, and (d) r = R? *I don't need answer a. * hint: Did you construct a Gaussian sphere through any given radial point, concentric with the actual sphere? What is the charge enclosed by the Gaussian sphere? (Do you see that with a variable density, integration is required?) Did you then apply Gauss' law to get the field?
An insulating sphere with radius R contains a total non-uniform charge (i.e. Hydrogen atom) Q such that its volume charge density is p = 3B / r3/2 (division) where is a constant and r is the distance from the center of the sphere. What is electric field at any point inside the sphere?Continuation of the problem can be seen in the images. Thanks!
An infinitely long cylindrical shell, with uniform charge density p, has an inner radius of X and an outer radius of Y. An infinitely long line consisting of uniform charge density L is located right at the axis of the cylindrical shell. A. With the help of a diagram of a Gaussian surface, find the equation for the electric feild strength in a region where r>X and r<Y. B. With the help of a diagram of a Gaussian surface, find the equation for the electric feild strength in a region where r<X C. A charge, Q, with a mass of M is released at rest. The charge is released at a radius of 10Y. Derive an equation for the direction and magnitude of the initial acceleration of the charge
If a solid insulating sphere of radius 50 cm carries a total charge of 150 µC uniformly distributedthroughout its volume, what is its a) volume charge density? What is the magnitude of the electricfield at b) 10 cm and c) 65 cm from the center of the sphere.
A sphere of radius 2m is made of a non-conducting material that has a uniformvolume charge density ρ = 2.655 × 10−10C/m3. A spherical cavity of radius 1m isthen carved out from the sphere. As measured from the center of the large sphere,the center of the spherical cavity is at the position ~r = cos300ˆi + sin300ˆj.Find the electric field at a point P within the cavity. As measured from the centerof the cavity, the point P is at the position ~r′ = 0.5cos600ˆi + 0.5sin600ˆj.1
A solid conducting sphere of radius R has a uniform charge distribution, with a density = Ps * r / R where Ps is a constant and r the distance from the center of the sphere. Prove a) the total charge on the sphere is Q = πPsR ^ 3 b) the electric field of the sphere is given by E = (1 / 4πε0) * (Q / R ^ 4) * (r ^ 2)
A small conducting ball that carries a charge of 30.0 nC islocated at the center of an electrically neutral hollow conductingsphere of inner radius 100 mm and outer radius 150 mm. Whatis the surface charge density (a) on the sphere’s inner surface and(b) on its outer surface? (c) Determine the electric field as a functionof r, the radial distance from the center of the sphere
A uniformly charged sphere, made of insulating material, has a radius a and a total charge of 2Q. The sphere is concentric with a conducting spherical shell that has inner radius b and outer radius c. (see figure). The shell has charge of 2Q as wella) Using Gauss’s Law, and with adequate explanation, derive an expression for the electric field inside the sphere, r<a)b) Using Gauss’s Law, and with adequate explanation derive an expression for the electric field outside the sphere, but before the inner surface of the shell a<r<b)c) Using Gauss’s Law, and with adequate explanation derive an expression for the electric field inside the metalshell, b<r<cd) Using Gauss’s Law, and with adequate explanation derive an expression for the electric field outside the metal shell, r>c,e) What is the charge on the inner surface of the metal shell? explain