In: Physics
A spherical ball of charge has radius R and total charge Q. The electric field strength inside the ball (r?R) is E(r)= Emax=(r4/R4)
A) What is Emax in terms of Q and R? Express your answer in terms of the variables Q, R, and appropriate constants.
B) Find an expression for the volume charge density ?(r) inside the ball as a function of r. Express your answer in terms of the variables Q, r, R, and appropriate constants.
F = q/e0
where e is the permittivity (epsilon_0 in your
notation)
The flux though a surface is given by the surface integral of the
dot product of the electric field and the unit normal to the
surface. In this case, the electric field is spherically symmetric,
so if we use a spherical Gaussian surface, this integral reduces
to
F = A * E
where E is the magnitude of the electric field, and A is the area
of the surface.
Putting these two equations together gives:
E = q/(e0 * A)
which expresses the electric field as a function of the charge
inside a closed surface and the area of that
surface.
The
surface area of a sphere
with radius r (our Gaussian surface) is:
A(r) = 4*pi*r^2,
so we can write:
E(r) = q/4*pi*r^2
For question 1, take the closed surface to be a sphere with radius
r > r_b. The total charge contained in this surface is simply
the total charge on the charged sphere, q. Note that the resulting
field is the same as one would get if all the charge on the charged
ball were concentrated at a point at the center of the ball (i.e.,
the equation for the field is the same as that of a point
charge).
To get q in terms of the parameters given in this problem, you need
to recognize that the total charge is simply the volumetric charge
density multiplied by the volume:
q = ((4/3)* pi * (r_b)^3) * rho
so, the electric field at a distance r > r_b is
then:
E(r) = [((4/3)* pi * (r_b)^3) * rho]/(e0 *
4*pi*r^2)
E(r) = (rho * (r_b)^3)/(3 * e0 * r^2)
At a point inside the charged ball (r < r_b) the amount of
charge enclosed by the spherical Gaussian surface will be less than
the total charge of the ball. In this case,
q(r) = rho * (4/3)*pi*(r^3)
so the field inide the charged ball is given by:
E(r) = (rho * (4/3)*pi*(r^3))/(4 * pi * r^2 * e0)
E(r) = (rho * r)/(3 * e0)
Note that when r = r_b, the two cases give the same answer for the
electric field at the surface of the charged ball.