Question

A small hollow conducting sphere of radius $r_1$ is given a charge Q. It is surrounded by a concentric conducting spherical shell of inner radius $r_2$ and outer radius $r_3$, having charge $-3q$. If a point charge 2q were kept at the centre, find the electric flux through a concentric spherical Gaussian surface of radius x for (1) $x < r_1$, and (2) $r_1 < x < r_2$.

Hint:

When painstakingly calculating enclosed charge for Gauss's law, you must absolutely include every single charge residing physically inside the drawn boundary—point charges, surface charges, and volume charges alike.
Flux strictly cares exclusively about the quantity of enclosed charge, not its detailed spatial arrangement, provided the surface is completely closed.

Answer 1

Concept:
• Gauss's law universally and strictly dictates that the total electric flux piercing entirely through any closed mathematical surface is precisely equal to the dynamically enclosed net charge divided directly by the permittivity of free space $\epsilon_0$.

• To successfully apply this law, one must carefully systematically define the exact imaginary Gaussian surface and meticulously sum all internal charges existing strictly within that spatial boundary.

• The spherical symmetry of concentric conducting shells makes applying a spherical Gaussian surface the optimal geometric choice.

Step 1: Evaluate Electric Flux for region (1) $x < r_1$
We systematically apply Gauss's integral law mathematically stated as $\Phi = \frac{Q_{enc}}{\epsilon_0}$ for distinct radial geometric regions.
For this innermost region strictly defined by radial distance $x < r_1$, the arbitrarily chosen spherical Gaussian surface perfectly encloses only the central space inside the first hollow sphere.
The only electrical entity existing in this deep central space is the centrally placed point charge.
Thus, the definitively enclosed net charge is mathematically calculated as exactly $Q_{enc} = 2q$.
Applying Gauss's law directly, the total electric flux piercing outward through this inner surface is firmly computed as:
\[ \Phi_{(x < r_1)} = \frac{2q}{\epsilon_0} \]

Step 2: Evaluate Electric Flux for region (2) $r_1 < x < r_2$
We logically shift our analytical focus to the intermediate spatial region located squarely between the inner conducting sphere and the outer conducting shell.
A spherical Gaussian surface meticulously drawn at any radius $x$ strictly bounded by $r_1 < x < r_2$ physically encompasses everything inside the inner sphere's radius.
This overarching Gaussian surface successfully surrounds both the exact central point charge ($2q$) and the entire primary inner conducting sphere, which inherently carries a net total charge of $Q$.
The newly enclosed net total charge strictly aggregates by scalar addition to:
\[ Q_{enc} = 2q + Q \]
Invoking Gauss's law once again using this updated enclosed charge value, the corresponding electric flux physically passing entirely through this intermediate surface is:
\[ \Phi_{(r_1 < x < r_2)} = \frac{2q + Q}{\epsilon_0} \]
This comprehensively concludes the explicit flux calculations for both highly specific interior regions.