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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=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 integral 3B denc = 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/ (Qr /R

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=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 integral 3B denc = 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/ (Qr /R

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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 =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 integral 3B Jenc = J. p3/2 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/ (Qr /R
Transcribed Image Text: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 =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 integral 3B Jenc = J. p3/2 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/ (Qr /R
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 p= 3/2 where Bis 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 dq p= dV Based on the given problem, we can also say that dqenc p= 3B 3/2 dV Let us first solve for B. Our enclosed charge would have limits from 0 to Q. Thenr would have the limits 0 to R. Thus, the equation above becomes 3B o p3/2 where dV is the infinitesimal volume.
Transcribed Image Text: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 p= 3/2 where Bis 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 dq p= dV Based on the given problem, we can also say that dqenc p= 3B 3/2 dV Let us first solve for B. Our enclosed charge would have limits from 0 to Q. Thenr would have the limits 0 to R. Thus, the equation above becomes 3B o p3/2 where dV is the infinitesimal volume.
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