The electrostatic potential in a charged spherical region of radius r varies as V=ar³+b, where a and b are constants. The total charge in the sphere of unit radius is α× π a∈₀. The value of α is ____ .
(permittivity of vacuum is ε₀ )

Solution

Physics sir · 4 steps

Get the field from the potential, then let Gauss's law count the charge inside the unit sphere.

1. Field from potential. The potential depends only on r, so the field is radial and E = -dV/dr = -3ar². The constant b drops out.
2. Apply Gauss's law. Take a spherical Gaussian surface of radius r. The field has the same size everywhere on it and points along the normal, so the flux is E × 4π r².

   Q/ε₀ = E × 4π r² = (-3ar²)(4π r²) = -12π a r⁴

3. Put r = 1. At r = 1 this gives Q = -12π aε₀. Comparing with α × π aε₀ gives α = -12.
4. So. α = -12, which is option (a).

The idea to keep. When V depends only on r, find E = -dV/dr first. Then the flux through a sphere, E × 4π r², gives the enclosed charge directly, with no integration needed.

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