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An insulating sphere with radius R contains a total non-uniform charge (e. Hydrogen atom) Q such that its volume charge density is 38 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 symmesric Gaussian surface, which is the surface of a sphere, then evaluate the dot product to abtain (Equation 1) The issue however is how much charge does the Gaussian surface encioses? 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 Based on the given problem, we can also say that daenc 38 dV 3/2

An insulating sphere with radius R contains a total non-uniform charge (e. Hydrogen atom) Q such that its volume charge density is 38 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 symmesric Gaussian surface, which is the surface of a sphere, then evaluate the dot product to abtain (Equation 1) The issue however is how much charge does the Gaussian surface encioses? 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 Based on the given problem, we can also say that daenc 38 dV 3/2

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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 38 312 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 obcain (Equation 1) The issue however is how much charge does the Gaussian surface encoses? 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 dgenc p= 38 dV
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 38 312 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 obcain (Equation 1) The issue however is how much charge does the Gaussian surface encoses? 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 dgenc p= 38 dV
Let us first solve for B. Our enclosed charge would have limits from O to Q. Then r would have the limits 0 to R. Thus. the equation above becomes R 3B AP- Q= ), 312 where aV is the infinizesimal volume. By evaluating the integral and simpifying. we obrain 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 spply the same definition of volume charge density. to obtain the integral 3B dv 3/2 denc the difference is now, the limits of r will be from 0 to r. Evaluating the integral and simplifying, we obtain Cenc = By substitution to Equation 1 above, then using A = and simplifying, we obtain E= (1/ KQr /R
Transcribed Image Text:Let us first solve for B. Our enclosed charge would have limits from O to Q. Then r would have the limits 0 to R. Thus. the equation above becomes R 3B AP- Q= ), 312 where aV is the infinizesimal volume. By evaluating the integral and simpifying. we obrain 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 spply the same definition of volume charge density. to obtain the integral 3B dv 3/2 denc the difference is now, the limits of r will be from 0 to r. Evaluating the integral and simplifying, we obtain Cenc = By substitution to Equation 1 above, then using A = and simplifying, we obtain E= (1/ KQr /R
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