Gauss\'s law. A solid sphere of radius 0.4m has a total positive charge of 26.8

Question

Gauss's law. A solid sphere of radius 0.4m has a total positive charge of 26.8 µC uniformly distributed throughout its volume. We want you to calculate the magnitude of the electric field at the following distances. To do this we need to remember the steps in Gauss's Law. First we recognize that we are dealing with spherical symmetry, so we can write down the left-hand-side of the Gauss's Law equation. After that, we simply need to determine the amount of charge enclose by a sphere with a radius equal to our position of interest. The electric field will equal the charge enclosed divided by epsilon, and then divided by the left-hand-side of the Gauss's Law equation.

A solid sphere of radius 0.4 m has a total positive charge of 26.8 C uniformly distributed throughout its volume. we want you to calculate the magnitude of the electric field at the following distances. To do this we need to remember the steps in Gauss's Law. First we recognize that we are dealing with spherical symmetry, so we can write down the left-hand-side of the Gauss's Law equation. After that, we simply need to determine the amount of charge enclose by a sphere with a radius equal to our position of interest. The electric field will equal the charge enclosed divided by epsilon, and then divided by the left-hand-side of the Gauss's Law equation. So let's solve for the following distances: (a) At the center of the sphere (should be easy, how much charge is enclose by a sphere of radius zero?) kN/C (b) 15.5 cm from the center of the sphere kN/C (c) At the surface of the sphere (now how much charge is enclosed?) kN/C (d) 69.5 cm from the center of the sphere kN/C

Explanation / Answer

radius, r = 0.4m
q =  26.8 µC

(a)
As No Charge is Enclosed by a Sphere of radius 0.
Therefore, E = 0 at Center of the Sphere.

(b)
Now , At Distance 15.5 cm = 0.155 m from the Center of Sphere.
There is uniform distribution of Charge throughout its volume -

Charge in Sphere of Vol, 4/3* * R^3 = 26.8 µC
Charge in Sphere of Vol, 4/3* * r^3 = 26.8 µC * (4/3* * r^3 )/ (4/3* * R^3)
Charge in Sphere of Vol, 4/3* * r^3 = 26.8 µC * (r/R)^3

Electric Feld is Given by, E = Q/(4*r^2)
E = 26.8 µC * (r/R)^3 / (4r^2)
E = 26.8 * 10^-6 * r / (4R^3)
E = (26.8 * 10^-6 * 0.155) / (4*3.14*8.85 × 10^-12 * 0.4^3)
E = 5.84 * 10^5 N/C
Electric Field At Distance 0.155 m from the Center of Sphere, E = 5.84 * 10^5 N/C

(c)
At the Surface of the Sphere, All the Charge is Enclosed,
Therefore,
E = Q/4*R^2
E = 26.8 * 10^-6 /  (4*3.14*8.85 × 10^-12 * 0.4^2) N/C
E =  1.506 * 10^6 N/C

Electric Field At the Surface of the Sphere, All the Charge is Enclosed, E =  1.506 * 10^6 N/C

(D)
For a Distance 69.5 cm from the Sphere , the Sphere will act as Point charge and the Value of Electric Field will be same as at the Surface.Therefore,
Electric Field at Distance 69.5 cm from the Sphere , E =   1.506 * 10^6 N/C

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