Question

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(r^(4)/R^(4)).

1) What is Emax in terms of Q and R?
2) Find an expression for the volume charge density ?(r) inside the ball as a function of r.

Answer 1

To solve this question, use Gauss' Law, which states that the total electric flux (F) through a closed surface (called a "Gaussian surface") that contains within it a charge, q, is given by:

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 bal

Answer 2

A spherical ball of charge has radius R$R$ and total charge Q$Q$. The electric field strength inside the ball (r≤R)$(r\le R)$ is E(r)=r4Emax/R4$E(r)=r^4{E}_{\mathrm{m}\mathrm{a}\mathrm{x}}/R^4$.A spherical ball of charge has radius R$R$ and total charge Q$Q$. The electric field strength inside the ball (r≤R)$(r\le R)$ is E(r)=r4Emax/R4$E(r)=r^4{E}_{\mathrm{m}\mathrm{a}\mathrm{x}}/R^4$.A spherical ball of charge has radius R$R$ and total charge Q$Q$. The electric field strength inside the ball (r≤R)$(r\le R)$ is E(r)=r4Emax/R4$E(r)=r^4{E}_{\mathrm{m}\mathrm{a}\mathrm{x}}/R^4$.A spherical ball of charge has radius A spherical ball of charge has radius R$R$ and total charge Q$Q$. The electric field strength inside the ball (r≤R)$(r\le R)$ is E(r)=r4Emax/R4$E(r)=r^4{E}_{\mathrm{m}\mathrm{a}\mathrm{x}}/R^4$.A spherical ball of charge has radius R$R$ and total charge Q$Q$. The electric field strength inside the ball (r≤R)$(r\le R)$ is E(r)=r4Emax/R4$E(r)=r^4{E}_{\mathrm{m}\mathrm{a}\mathrm{x}}/R^4$.A spherical ball of charge has radius R$R$ and total charge Q$Q$. The electric field strength inside the ball (r≤R)$(r\le R)$ is E(r)=r4Emax/R4$E(r)=r^4{E}_{\mathrm{m}\mathrm{a}\mathrm{x}}/R^4$.A spherical ball of charge has radius R$R$ and total charge Q$Q$. The electric field strength inside the ball (r≤R)$(r\le R)$ is E(r)=r4Emax/R4$E(r)=r^4{E}_{\mathrm{m}\mathrm{a}\mathrm{x}}/R^4$.

R$R$ and total charge Q$Q$. The electric field strength inside the ball (r≤R)$(r\le R)$ is E(r)=r4Emax/R4$E(r)=r^4{E}_{\mathrm{m}\mathrm{a}\mathrm{x}}/R^4$.