Problem 54
Question
A solid, nonconducting sphere of radius \(a\) has total charge \(Q\) and a uniform charge distribution. Using Gauss's Law, determine the electric field (as a vector) in the regions \(ra\) in terms of \(Q\).
Step-by-Step Solution
Verified Answer
Answer: The electric field E(r) is as follows:
- Inside the sphere (r < a): E(r) = (Q*r) / (4π*ε₀*a^3) * rˆ
- Outside the sphere (r > a): E(r) = Q / (4π*ε₀*r^2) * rˆ
1Step 1: Determine the Gaussian surface
Choose the Gaussian surface as a concentric sphere with radius r, where r is the distance from the center of the sphere. Using symmetry, it is evident that the electric field will have only a radial component.
2Step 2: Applying Gauss's Law - Inside the sphere (r < a)
Let's consider the spherical Gaussian surface with radius r inside the sphere (r < a). As the electric field (E) and the area vector (dA) are both radially pointing outward (same direction), the angle between them (θ) is 0 degrees. Therefore, the dot product of E and dA in Gauss's Law is |E||dA|cosθ which simplifies to |E||dA|.
To apply Gauss's Law, we need to determine the charge enclosed by the Gaussian surface of radius r. Since the sphere has a uniform charge density, we can find the charge density (ρ) as:
ρ = Q / (4/3*π*a^3)
The volume enclosed by the Gaussian surface is:
V = (4/3)*π*r^3
Hence, the enclosed charge (q) is:
q = ρ*V = Q*(r^3) / (a^3)
Now, apply Gauss's Law to the Gaussian surface of radius r:
∮E⋅dA = q / ε₀
|E|∮dA = q / ε₀
|E|(4πr^2) = (Q*r^3)/(a^3*ε₀)
Solving for E, we get for r < a:
E = (Q*r)/(4π*ε₀*a^3)
3Step 3: Applying Gauss's Law - Outside the sphere (r > a)
Now let's consider the spherical Gaussian surface with radius r outside the sphere (r > a). For this case, the angle (θ) between E and dA is still 0. The Gaussian surface encloses the whole sphere, hence enclosed charge (q) remains Q.
Applying Gauss's Law to the Gaussian surface of radius r:
∮E⋅dA = q / ε₀
|E|∮dA = Q / ε₀
|E|(4πr^2) = Q/ε₀
Solving for E, we get for r > a:
E = Q / (4π*ε₀*r^2)
4Step 4: Writing the electric field as a vector
Since the electric field has only a radial component, we can write it in spherical coordinates as:
E(r) = Er * rˆ = |E| * rˆ
Combining our results for r < a and r > a, we get:
Electric field E(r) =
- For r < a: (Q*r) / (4π*ε₀*a^3) * rˆ
- For r > a: Q / (4π*ε₀*r^2) * rˆ
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