Electric Fields

Question

A solid sphere with a radius of 1 cm bears a uniform volumetric charge density of 300 mC/m3. It is placed at the origin. A thin spherical shell with a radius of 2 cm and a uniform surface charge density of 1 mC/m2 is placed on the x-axis at x=3 km. There is a point somewhere between them where E = 0. Find that location.

Answer 1

I will set this up for you and work through all of the equations, and let you crunch out the final number at the end.  

 

Set up Gauss' law around each sphere.

 

Sphere 1:

 

A₁(r) E₁(r) = Q₁/ε₀

 

a = radius of sphere = 1 cm

 

4πr² E₁(r) = (4/3)πa³ρ/ε₀  [only true for r > a]

 

E₁(r) = ρa³ / (3ε₀r²)

 

Sphere 2:

 

b = radius = 2 cm

x = coordinate on x-axis of center of sphere = 3 km

 

Set up Gauss' law around the center of sphere 2, with a radius of (x - r) for the Gaussian surface.  Find E₂, and set the magnitudes of E₁ and E₂ equal, since the 2 fields must be equal and opposite at the point where the 2 Gaussian surfaces touch, which is r from the center of sphere 1, which is the same as x - r from the center of the sphere 2.  Then you can solve for r.

 

Gauss' law around sphere 2:

 

A₂(r) E₂(r) = Q₂/ε₀

 

4π(x - r)²E₂(r) = 4πb²σ/ε₀

 

E₂(r) = b²σ / [ε₀(x - r)²]

 

Set E₁(r) = E₂(r) and solve for r.

 

ρa³ / (3r²) = b²σ / (x - r)²

 

ρ = volume charge density of sphere 1

σ = surface charge density of sphere 2

a = 1 cm

b = 2 cm

x = 3 km

 

At this point, the physics is done, and it is just an algebra problem.  Plug in the numbers and solve for r, being careful with the units.  You should be able to finish from here.