Question

find the electric field at point p due outside the spherical charge distributions of radius 9 nm. where point p is located at a distance of 3 nm from the sphere having density of 2000000 C\nm*3.

Answer 1

Answer:

$E=6.10\cdot 10^{36} \ N/c$

Explanation:

Gauss' Law

The total of the electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity.

Expressed as an equation:

$\displaystyle \Phi =\frac{Q}{\epsilon}$

Where Q is the enclosed charge and \epsilon is the permittivity.

The electric field of a conducting sphere with charge Q can be obtained by an application of Gauss' law.

Considering a Gaussian surface in the form of a sphere at radius r > R, the electric field has the same magnitude at every point of the surface and is directed outward. The electric flux is then just the electric field times the area of the spherical surface:

$\displaystyle \Phi =E.A$

The surface of a sphere is

$A=4\pi r^2$

Thus:

$\displaystyle \Phi =E.(4\pi r^2)$

Applying Gauss' Law:

$\displaystyle E.(4\pi r^2) =\frac{Q}{\epsilon}$

Solving for E:

$\displaystyle E =\frac{Q}{4\pi\epsilon r^2}$

The charge Q is the density by the volume of the sphere:

$Q=\rho.V$

The volume of the sphere is:

$\displaystyle V=\frac{4}{3}\pi R^3$

Where R is the radius of the sphere:

$\displaystyle V=\frac{4}{3}\pi (9\eta m)^3$

$V=3053.63\ \eta m^3$

The total enclosed charge is:

$Q=2000000\cdot 3053.63$

$Q=6.11\cdot 10^{9}\ c$

The electric field at $r=3\ \eta m$ is

$\displaystyle E =\frac{6.11\cdot 10^{9}}{4\pi\cdot 8.85\cdot 10^{-12} (3\cdot 10^{-9})^2}$

$\boxed{E=6.10\cdot 10^{36} \ N/c}$