The electrostatic potential inside a charged spherical ball is given by φ=ar²+b, where r is the distance from the centre and a, b are constants. Then the charge density inside the ball is

Question

Accepted Answer

(4) -6aε₀. Get the field first: E=-(dφ)/(dr)=-2ar, directed radially. Apply Gauss's law to a sphere of radius r: q_enc=ε₀ E 4π r²=ε₀(-2ar)(4π r²)=-8π aε₀ r³. The density is ρ=1/(4π r²)(dq_enc)/(dr)=(-24π aε₀ r²)/(4π r²)=-6aε₀. (The same result follows at once from Poisson's equation, ρ=-ε₀
abla²φ=-ε₀·6a.) The density is uniform, independent of r.

The electrostatic potential inside a charged spherical ball is given by φ=ar²+b, where r is the distance from the centre and a, b are constants. Then the charge density inside the ball is

Step-by-step solution

Why the other options are wrong