Refer to the diagram not to scale which shows a sphere of ou

Refer to the diagram (not to scale), which shows a sphere (of outer radius R) that is made of solid glass except for a spherical cavity at its center (of radius R/3), which is empty. The sphere carries a total net charge of -Q that is uniformly distributed throughout the glass (but no charge resides in the empty central cavity). In terms of R, Q, k (or epsilon_0), find an expression for the electric field (magnitude and direction) - as a function of distance r from the sphere\'s center-for the following regions: 0

Solution

For the given problem, we will make use of the Gauss\'s law which gives us the relation between line integral of electric field along a path, around a charge distribution and the net charge enclosed within that path.

We will consider suitable Gaussian surfaces for each of the parts and the resolve.

Part a.) For 0 <= r < R/3

Let us assume a gaussian surface with radius r satisfying the above condition.

Now by Gauss\'s law we can write:

E.ds = Q_enclosed / _o

Now for the surface chosen, it lies inside the cavity and hence has no charge enclosed within it. Therefore the electric field for the given region would be zero.

b.) For R/3 < r < R

Let us assume a Gaussian surface satisfying the above condition. By Gauss\'s law, we get:

E.ds = Q_enclosed / _o

Now, Q_enclosed = Charge density x volume

or, Qenc = [Q / (4/3)(R³ - R³/27)] x [(4/3)(r³ - R³/27)]

or, Qenc = -Q(r³ - R³/27) / (R³ - R³/27)

Putting into the relation for the Gauss\'s law we get:

E (4r²) = -Q(r³ - R³/27) / _o(R³ - R³/27)

or, E = -Q(r³ - R³/27) / 4r²_o(R³ - R³/27) is the required expression for electric field. The negative sign shows that the direction of the electric field is towards the centre of the sphere.

Part c.) For r > R

Let us assume a Gaussian surface of radius r and concentric with the given sphere.

By Gauss\'s law we have:

E.ds = Q_enclosed / _o

Charge enclosed = -Q

Hence, E(4r²) = -Q / _o

or, E = -Q / 4r²_o is the required expression for the electric field and is directed inwards.