a positive point charge q sits at the center of a hollow spherical shell. The shell, with radius R and negligible thickness, has net charge -2q. find an expression for the for the electric field strength (a) inside the sphere, r<R (b) outside the sphere, r>R. in what direction does the electric field point in each case.

To find the electric field strength inside and outside the hollow spherical shell, we can apply Gauss's Law. Gauss's Law states that the flux of the electric field through a closed surface is proportional to the charge enclosed by that surface.

Let's consider a Gaussian surface in the form of a sphere of radius r, where r is less than R.

(a) Inside the sphere (r < R):
Inside the sphere, the Gaussian surface encloses both the positive point charge q at the center and the negative charge -2q on the shell. Since the shell has a net charge of -2q, the total charge enclosed by the Gaussian surface is -q (charge of the point charge at the center).

According to Gauss's Law, the electric flux through a closed surface is proportional to the charge enclosed. Therefore, the electric field inside the sphere due to the charge on the shell will be canceled out by the charge at the center. So, the net electric field inside the sphere is zero.

In summary, the expression for the electric field strength inside the sphere (r < R) is E = 0, and the direction of the electric field is undefined because it is zero.

(b) Outside the sphere (r > R):
Outside the sphere, the Gaussian surface does not enclose any charge, as the point charge q and the shell are not enclosed within the surface. Therefore, the total charge enclosed by the Gaussian surface is zero, and according to Gauss's Law, the electric flux will also be zero.

Since the electric flux is zero, the net electric field outside the sphere (r > R) is also zero.

In summary, the expression for the electric field strength outside the sphere (r > R) is E = 0, and again, the direction of the electric field is undefined because it is zero.

To summarize:
(a) Inside the sphere (r < R): Electric field strength is zero, and the direction is undefined.
(b) Outside the sphere (r > R): Electric field strength is zero, and the direction is undefined.

(a) Inside the sphere, r < R:

To find the electric field inside the sphere, we can use Gauss's Law. Since the charge distribution is spherically symmetric, we can imagine a Gaussian surface in the form of a sphere with radius r, where r is the distance from the center.

By Gauss's Law, the total electric flux passing through the Gaussian surface is given by

Φ = (Electric field strength inside the sphere) * (Surface area of the Gaussian surface).

Since there is no charge enclosed by the Gaussian surface, the net electric flux through the Gaussian surface is zero.

Therefore,

Φ = 0.

The electric field is radial and points outward, so it is in the same direction as the surface normal vector.

The surface area of the Gaussian surface is given by 4πr^2.

Therefore, we have:

Φ = (Electric field strength inside the sphere) * 4πr^2 = 0.

Solving for the electric field strength inside the sphere:

Electric field strength inside the sphere = 0

(b) Outside the sphere, r > R:

To find the electric field outside the sphere, we can use Gauss's Law again. We can imagine a Gaussian surface in the form of a sphere with radius r, where r is the distance from the center.

By Gauss's Law, the total electric flux passing through the Gaussian surface is given by

Φ = (Electric field strength outside the sphere) * (Surface area of the Gaussian surface).

Since there is charge inside the Gaussian surface, the net electric flux through the Gaussian surface is given by

Φ = (Electric field strength outside the sphere) * 4πr^2.

The total charge enclosed by the Gaussian surface is q (positive point charge) + (-2q) (charge on the shell).

Therefore, by Gauss's Law,

Φ = (q + (-2q))/(ε0),

where ε0 is the permittivity of free space.

Simplifying,

Φ = -q/ε0.

Setting this equal to the electric flux expression,

(Electric field strength outside the sphere) * 4πr^2 = -q/ε0.

Solving for the electric field strength outside the sphere:

Electric field strength outside the sphere = -q/(4πε0r^2).

The electric field is radial and points radially inward due to the negative charge distribution on the shell.

Have you tried using Gauss' law?

It is well suited to problems like this.

http://hyperphysics.phy-astr.gsu.edu/HBASE/electric/elesph.html