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:
(I) the electric flux through a concentric spherical Gaussian surface of radius \(x\) for (1) \(x < r_1\), and (2) \(r_1 < x < r_2\).
(II) electric field at a point distant \(x\) from the centre for (1) \(x > r_3\), and (2) \(r_1 < x < r_2\).
(III) surface charge density on the inner surface of (1) sphere, and (2) shell.

Hint:

For concentric conducting shells, remember that the charge induced on an inner surface is always exactly equal in magnitude and opposite in sign to the total net charge residing everywhere inside that specific boundary.

Answer 1

Concept: Gauss's law states that the total electric flux \(\Phi_E\) passing through any closed boundary surface is equal to exactly \(\frac{1}{\varepsilon_0}\) times the total algebraic net charge enclosed within that volume: \[ \Phi_E = \oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enclosed}}}{\varepsilon_0} \] For spherically symmetric systems, the magnitude of the radial electric field at a distance \(x\) from the center is given directly by: \[ E(x) = \frac{Q_{\text{enclosed}}}{4\pi\varepsilon_0 x^2} \] By using electrostatics of conductors, charges inside conducting bodies rearrange themselves via induction to maintain an interior electric field of exactly zero.

Step 1: Solving Part (I) - Electric Flux through a Gaussian Surface of radius \(x\).
Calculate the net enclosed charge within each given radial limit.
- Case (1) for \(x < r_1\): The Gaussian surface is situated inside the hollow interior space of the first inner conducting sphere. The only charge enclosed inside this sphere is the point charge sitting exactly at the absolute center, which is \(Q_{\text{enclosed}} = 2q\). Therefore, applying Gauss's Law: \[ \Phi_E = \frac{2q}{\varepsilon_0} \] - Case (2) for \(r_1 < x < r_2\): The Gaussian surface lies in the open region between the outer perimeter of the inner sphere and the inner boundary of the external shell. The enclosed charges inside this boundary include both the central point charge \(2q\) and the total charge distributed over the small inner conducting sphere, which is \(Q\). Thus, \(Q_{\text{enclosed}} = Q + 2q\). The total flux is: \[ \Phi_E = \frac{Q + 2q}{\varepsilon_0} \]

Step 2: Solving Part (II) - Electric Field at a distance \(x\) from the center.
Apply the spherical field equation by substituting total enclosed charge within the boundaries.
- Case (1) for \(x > r_3\): The point lies entirely outside the whole assembly. The net combined charge enclosed by a giant outer spherical boundary contains the central point charge (\(2q\)), the inner sphere charge (\(Q\)), and the outer large shell charge (\(-3q\)): \[ Q_{\text{enclosed, total}} = 2q + Q + (-3q) = Q - q \] Using the spherical configuration formula for electric fields: \[ E = \frac{1}{4\pi\varepsilon_0}\frac{Q - q}{x^2} \] - Case (2) for \(r_1 < x < r_2\): As established in Step 1, Case 2, the total net charge contained inside this intermediate radial zone is \(Q_{\text{enclosed}} = Q + 2q\). Therefore, the magnitude of the field is: \[ E = \frac{1}{4\pi\varepsilon_0}\frac{Q + 2q}{x^2} \]

Step 3: Solving Part (III) - Surface Charge Densities \(\sigma\).
Determine induced charges on conducting surfaces to find charge per unit area.
- (1) Inner surface of the inner sphere (radius \(r_1\)): To keep the electric field inside the solid bulk conducting material of the sphere equal to zero, an equal and opposite charge must be induced on its inner surface to completely screen out the central point charge \(2q\). Thus, the induced charge on the inner surface of the sphere is \(q_{\text{inner sphere}} = -2q\). Since surface area is \(4\pi r_1^2\), the surface charge density is: \[ \sigma_{\text{inner sphere}} = \frac{-2q}{4\pi r_1^2} \] - (2) Inner surface of the shell (radius \(r_2\)): The net charge on the entire inner sphere system combined is the sum of its outer surface, inner surface, and central core: \(2q + Q\). To completely screen this net charge and maintain a zero electric field inside the bulk material of the outer conducting shell, an equal and opposite charge must be induced on the inner surface of the shell at radius \(r_2\). Thus, the induced charge here is \(q_{\text{inner shell}} = -(Q + 2q)\). The surface charge density is: \[ \sigma_{\text{inner shell}} = \frac{-(Q + 2q)}{4\pi r_2^2} \]