Using Gauss's law



charged sphere

We can use Gauss's law to calculate the electric field of a sphere with a spherically symmetric charge. For simplicity, we will use a constant charge density.



Gauss law equation

We begin with the general Gauss's law, relating electric flux to a closed integral, to the enclosed charge.



Gauss law equation

To find the electric field everywhere outside the sphere, We integrate over the outer surface area of the sphere.



Gauss law equation

Using the relationship that the total flux equals the charge enclosed over the permittivity constant, We can solve to the E field. The field points radially away from the spherically symmetric charge density.



charged sphere

We can use a similar method to find the electric field inside a uniform sphere of charge.



Gauss law equation

We use the definition of a constant volume charge density to write the equation for the charge enclosed inside a volume of radius r, inside the sphere.



Gauss law equation

Inserting this Qin, we can write the equation for the flux inside the sphere, and for the electric field inside the sphere.



Graph of sphere E field

Notice that the form of the equation is qualitatively different from the equation for the field outside the sphere. Inside the sphere, the field increases with r, while outside the sphere the field goes as 1/r2.



Conducting sphere



A conductor in electrostatic equilibrium has no current inside, that is, by definition, there are no charges moving. This means that the electric field inside the conductor must be zero, otherwise the electric field would cause charge carriers to move.



conducting sphere

Consider a conducting sphere. If the electric field is zero inside the sphere, then a Gaussian surface drawn just inside the surface of the sphere must have zero net flux. The Gaussian surface must have no charge enclosed. There must be no charge inside the conducting sphere in electrostatic equilibrium.



hollow conducting spheres

Consider a hollow conducting sphere with a point charge of +Q at the center and a charge of +3Q on the conductor. The point charge at the center will induce an equal and opposite charge on the inner surface of the conductor. That charge had to come from somewhere. It comes from the +3 charge on the conductor, leaving +4Q on the outer surface of the hollow sphere.



Sample problem



A hollow conducting sphere has an inner edge of radius a and an outer edge of radius b. The sphere has a charge of +2Q, and a point charge of +Q sits at the center of the sphere.



conducting sphere

Find the electric field for the regions:

regions