A solid, insulating sphere of radius a has a uniform charge density ρand a total charge Q . Conc

step 1 Hi, in the given problem there is a solid insulating sphere. First of all, this is the solid sph

A solid, insulating sphere of radius a has a uniform charge...

Key Concepts

Gauss's Law

This principle relates the electric flux through a closed surface to the charge enclosed by that surface. It is used to determine the electric field in regions of high symmetry by choosing an appropriate Gaussian surface and setting up the equation ?E·A = Q_enc/??. In problems with spherical symmetry, Gauss's law simplifies the calculation of the electric field by accounting for only the enclosed charge.

Spherical Symmetry

Spherical symmetry means that the system’s properties are uniform in all directions from the center point. This symmetry allows for the simplification of electric field calculations, as the field only depends on the radial distance from the center of the sphere. It enables the use of spherical Gaussian surfaces, which makes it easier to compute the electric flux and hence the electric field.

Charge Distribution

A uniform charge distribution implies that the charge density is constant over the volume of the material. In this context, the solid sphere has a uniform volume charge density, which leads to a predictable variation of the enclosed charge with radius. Understanding the charge distribution is crucial for applying Gauss's law correctly in regions of space both inside and outside the charged sphere.

Conducting Materials in Electrostatics

Conductors in electrostatic equilibrium have free charges that move until the electric field within the conducting material is zero. When placed in an electric field, such conductors will redistribute their surface charges, resulting in induced charges on their inner and outer surfaces. This concept is essential when analyzing systems where a conductor is influenced by external or internal charge distributions.

Induced Surface Charge

In response to an external electric field or an enclosed charge, a conductor will develop induced surface charges to cancel any internal electric field. The inner and outer surfaces of a conductor can accumulate different amounts of charge such that the net charge inside the conductor is zero, while the overall neutrality of the conductor is maintained. This key concept explains the distribution of surface charge density in problems involving conductors and ensures the electric field inside the conducting material remains zero.

Transcript

00:01 Hi, in the given problem there is a solid insulating sphere.

00:09 First of all, this is the solid sphere, insulating sphere whose radius is given as a.

00:20 Surrounding it there is a conducting shell which is concentrating with this.

00:28 So this is the concentric shell whose inner radius is a sorry b the inner radius is b and whose outer radius is c this is it becomes the outer radius which is c and this surrounding shell is conducting in nature then the charge density the volume charge density over the insulating solid sphere is given as row and charge over the same is given as q capital q in the first part of the problem we have to find the expressions for electric field at various points in this given arrangement.

01:54 So first of all we have to find the magnitude of electric field in the region at a distance are from the center of the solid sphere where this r is less than a means the observation point is somewhere here so the gaussian surface will be another spherical shell having a radius r for which first of all we will have to find the total charge enclosed within this gaussian surface to find it the charge is given by the volume charge density which is row multiplied by the volume of this gaussian surface means the volume of smaller sphere which is 4 by 3 pi r cube hence electric field at the surface of this gaussian surface will come out to be e is equal to 1 by 4 pi epsilon not into the charge and closed within this divided by its radius, square of its radius, which is r.

03:09 So finally it becomes 1 by 4 pi epsilon not into q, which is row, into 4 by 3 pi r cubed, divided by square of r here this 4 pi will get cancelled and this r square will be cancelled leaving just one r in the numerator so finally the answer for electric field at this desired range comes out to be e is equal to row multiplied by r divided by three epsilon knot and it becomes the answer the first answer of this first part of the problem as there are various parts in the problem now for the second part of the same problem the range of this distance is now given as between a and b means the observation point is now somewhere here at this point between a and b.

04:32 This is.

04:34 So gaussian surface will be having a radius equal to r means it will be lying between the solid sphere and the inner surface of the shell...

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