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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 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!

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
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Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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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!

where dV is the infinitesimal volume. By evaluating the integral and simplifying, we obtain the following Q= R for the limits from 0 to R Thus, B can then be expressed as B =QI 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 integral 3B „3/2 lenc = Ap- the difference is now, the limits of r will be from 0 to r. Evaluating the integral and simplifying, we obtain denc = By substitution to Equation 1 above, then using A = ,and simplifying, we obtain E = (1/ XQr /R
Transcribed Image Text:where dV is the infinitesimal volume. By evaluating the integral and simplifying, we obtain the following Q= R for the limits from 0 to R Thus, B can then be expressed as B =QI 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 integral 3B „3/2 lenc = Ap- the difference is now, the limits of r will be from 0 to r. Evaluating the integral and simplifying, we obtain denc = By substitution to Equation 1 above, then using A = ,and simplifying, we obtain E = (1/ XQr /R
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 fĒ - dà = 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 dqenc p= 3B dV p312 Let 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 becomes .R 3B dV p3/2 Q =
Transcribed Image Text: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 fĒ - dà = 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 dqenc p= 3B dV p312 Let 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 becomes .R 3B dV p3/2 Q =
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