Question

A conducting sphere with radius R carries total charge Q. What is the magnitude of the electric field at distances R/2 and 2R from the center of the sphere

Answer 1

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:

`\oint E \cdot dA = E\cdot A = (Q_(encl))/(\epsilon_0)`

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:

`E \cdot A = (0)/(\epsilon_0)\\\\\boxed{E = 0 (N)/(C)}`

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.

`E \cdot A = (Q_(encl))/(\epsilon_0)`

'A' is equivalent to the surface area of a sphere of radius '2R', or:

`A = 4\pi (2R)^2\\\\A = 4\pi (4R^2)\\\\A = 16\pi R^2`

Substituting this expression back into Gauss's Law:

`E \cdot 16\pi R^2 = (Q)/(\epsilon_0)\\\\E = (Q)/(16\pi R^2\epsilon_0)`

To simplify:

`E = (1)/(4\pi \epsilon_0 ) * (Q)/(4R^2)\\`

OR using k = 1/4πε₀:

`\boxed{E = (kQ)/(4R^2)}`