Hello!
Distance of R/2:
Since a conducting sphere is referenced in this situation, all of its charge will be distributed along its SURFACE. Therefore, there is NO enclosed at a distance of R/2 from the center.
Using Gauss's Law:
[tex]\oint E \cdot dA = E\cdot A = \frac{Q_{encl}}{\epsilon_0}[/tex]
E = Electric field strength (N/C)
A = Area of Gaussian surface (m²)
Q = Enclosed charge (C)
ε₀ = Permittivity of free space C²/Nm²)
If the enclosed charge is 0, then:
[tex]E \cdot A = \frac{0}{\epsilon_0}\\\\\boxed{E = 0 \frac{N}{C}}[/tex]
Distance of '2R':
We can once again use Gauss's Law to solve. This time, however, a surface of radius '2R' encloses ALL of the charge of the sphere.
[tex]E \cdot A = \frac{Q_{encl}}{\epsilon_0}[/tex]
'A' is equivalent to the surface area of a sphere of radius '2R', or:
[tex]A = 4\pi (2R)^2\\\\A = 4\pi (4R^2)\\\\A = 16\pi R^2[/tex]
Substituting this expression back into Gauss's Law:
[tex]E \cdot 16\pi R^2 = \frac{Q}{\epsilon_0}\\\\E = \frac{Q}{16\pi R^2\epsilon_0}[/tex]
To simplify:
[tex]E = \frac{1}{4\pi \epsilon_0 } * \frac{Q}{4R^2}\\[/tex]
OR using k = 1/4πε₀:
[tex]\boxed{E = \frac{kQ}{4R^2}}[/tex]
