A solid sphere of radius R has electric charge density given by 𝜌 = a/r , where a is constant. Find the electric field within the charge distribution as a function of r. Note: The volume element dV for a spherical shell of radius r and thickness dr is equal to 4𝜋r2dr. (Use the following as necessary: a, r, and 𝜀0. Consider that a is positive.)
Question
A solid sphere of radius R has electric charge density given by 𝜌 = a/r , where a is constant. Find the electric field within the charge distribution as a function of r. Note: The volume element dV for a spherical shell of radius r and thickness dr is equal to 4𝜋r2dr. (Use the following as necessary: a, r, and 𝜀0. Consider that a is positive.)
Solution
To find the electric field within the charge distribution, we first need to find the charge dq in a small shell of thickness dr at a distance r from the center. The volume of this shell is dV = 4πr²dr. Given the charge density ρ = a/r, the charge dq in this volume is given by dq = ρdV = a/r * 4πr²dr = 4πadr.
Next, we use Gauss's law, which states that the electric field E times the surface area of a Gaussian surface (in this case, a sphere of radius r) is equal to the enclosed charge divided by ε₀. The surface area of the sphere is 4πr², so we have E * 4πr² = ∫dq/ε₀ from 0 to r.
Substituting dq from above, we get E * 4πr² = ∫4πadr/ε₀ from 0 to r. The 4π terms cancel out, and we can integrate the right side with respect to r to get E * r² = ar²/2ε₀ from 0 to r.
Finally, we solve for E to get the electric field as a function of r: E = ar/2ε₀r² = a/2ε₀r. This is the electric field within the charge distribution as a function of r.
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