Using the results of Problem 75, we can find the electric field at any radius for any sphericall

Using the results of Problem 75, we can find the electric...

Key Concepts

Gauss's Law

Gauss's Law is a fundamental principle in electromagnetism that relates the electric flux through a closed surface to the charge enclosed by that surface. It is particularly useful for calculating electric fields where the symmetry of the charge distribution, such as spherical symmetry, allows the flux integral to be simplified.

Spherical Symmetry

Spherical symmetry means that the charge distribution is uniform in all directions around a central point, allowing the electric field to depend only on the radial distance from the center. This symmetry simplifies the application of Gauss's Law because the electric field has the same magnitude at every point on a spherical Gaussian surface.

Uniform Charge Distribution

A uniform charge distribution implies that the charge is spread evenly throughout the volume of the sphere. This results in a constant charge density, which is crucial for determining how much charge is enclosed by a Gaussian surface of a given radius, and it directly influences the form of the electric field both inside and outside the sphere.

Electric Field Behavior in Radial Directions

The electric field of a spherically symmetric charge distribution behaves differently depending on whether one is inside or outside the distribution. Outside the sphere, the field behaves as if the entire charge were concentrated at the center, leading to an inverse square law dependence with distance. Inside the sphere, the amount of enclosed charge changes with the radius, typically leading to a linear dependence on the distance from the center until the field reaches the surface value.

Transcript

00:02 Hi, in the given problem we have been given a solid metallic charged sphere like this.

00:20 Suppose it is having a total charge distributed everywhere within it, that is q, capital q.

00:55 Its radius is given as capital r and in the first part of the problem we have to find electric field at a point whose distance is smaller where this r is either equal to or greater than this capital r so suppose this observation point is here this is p at a distance small r so the gaussian surface passing through this observation point is in the form of a shell.

01:36 Now the charge, the total charge included within this gaussian surface is capital q plus suppose the observation point is p at a distance smaller hence the gaussian surface is in the form of a shell spherical shell of radius smaller hence using theorem which says the total electric flux linked with the gaussian surface that is equal to surface integral of electric field first of all and then also it is given as 1 upon epsilon times the total charge and closed by 8 now as the electric field lines are linked through this gaussian surface perpendicularly everywhere they are perpendicular.

03:10 So we can say the angle between area vector and electric field is 0 degree.

03:20 So here it will come out to be eds, cos 0 degree which later will become 1 is equal to 1 1 upon epsilon 0 times the charge and closed.

03:33 Electric field will be constant everywhere because it is at the same distance from the center of the solid sphere.

03:39 So that will come out to be constant remaining, leaving behind integration of ds only is equal to 1 upon epsilon not times the charge and closed.

03:50 So this ds means the area, surface area of the gaussian surface which is 4 pi r square is equal to 1 upon epsilon note times the charge and close.

04:00 So finally this electric field here will come out to be 1 upon 4 pi epsilon 0 into q by r square.

04:10 So it says electric field outside the sphere, the solid sphere, will be depending on the square of the distance inversely, which will be used in order to draw, in order to plot the curve for the electric field with the distance.

04:30 Now, in the second part of the problem, we have to find the same electric field but this time at a point which is within this solid sphere...

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