For a charged spherical ball, electrostatic potential inside the ball varies with r as V=2ar²+b.
Here, a and b are constant and r is the distance from the center. The volume charge density inside the ball is -λ aε. The value of λ is ____________.
ε = permittivity of the medium

Solution

Physics sir · 4 steps

Get E from V by differentiating, then put that E into Gauss's law for a sphere of radius r.

1. Field from potential. Inside the ball, E = -dV/dr = -d/dr(2ar² + b) = -4ar. The constant b drops out, so it plays no part in the charge density.
2. Gauss's law inside. Take a Gaussian sphere of radius r inside the ball. The charge it encloses is ρ · 4/3π r³ and the flux through it is E · 4π r².

   E · 4π r² = (ρ · 4/3π r³)/ε ⇒ E = (ρ r)/3ε

3. Match the two. Set (ρ r)/3ε = -4ar. Then ρ = -12aε, which is the same as -λ a ε with λ = 12.
4. So. The volume charge density is -12aε, so λ = 12.

The common slip: 4. Writing E = ρ r/ε and dropping the 3 that comes from the sphere's volume divided by its surface area gives 4, not 12.

The idea to keep. Potential given, charge wanted: differentiate to get E = -dV/dr, then use Gauss's law. Inside a uniformly charged sphere E = ρ r/(3ε), so do not drop that 3.

Answer from the NTA final answer key. Solution by PrepWiser.