Askiitians Tutor Team
Last Activity: 9 Months ago
A. Surface Charge Density on the Inner and Outer Surfaces of the Shell
We are given a spherical conducting shell with the following:
• Inner radius r1r_1,
• Outer radius r2r_2,
• Total charge QQ,
• A charge qq placed at the center of the shell.
(i) Surface Charge Density on the Inner Surface:
The shell is conducting, and the electric field inside the conducting material (i.e., between r1r_1 and r2r_2) must be zero. This implies that the charges within the conducting shell will redistribute in such a way that they cancel any internal electric fields.
• The charge qq placed at the center will induce a charge of −q-q on the inner surface of the conducting shell to neutralize the electric field inside the conductor.
• This means the inner surface of the shell will have a charge of −q-q on it.
The surface charge density σinner\sigma_{\text{inner}} on the inner surface is given by:
σinner=Charge on inner surfaceSurface area of inner surface=−q4πr12\sigma_{\text{inner}} = \frac{\text{Charge on inner surface}}{\text{Surface area of inner surface}} = \frac{-q}{4\pi r_1^2}
Thus, the surface charge density on the inner surface is:
σinner=−q4πr12\sigma_{\text{inner}} = \frac{-q}{4\pi r_1^2}
(ii) Surface Charge Density on the Outer Surface:
Since the shell is conducting and the total charge on the shell is QQ, the total charge on the shell must be distributed between the inner and outer surfaces. The charge on the inner surface is −q-q, so the total charge on the outer surface must be:
Qouter=Q−(−q)=Q+qQ_{\text{outer}} = Q - (-q) = Q + q
The surface charge density σouter\sigma_{\text{outer}} on the outer surface is given by:
σouter=Charge on outer surfaceSurface area of outer surface=Q+q4πr22\sigma_{\text{outer}} = \frac{\text{Charge on outer surface}}{\text{Surface area of outer surface}} = \frac{Q + q}{4\pi r_2^2}
Thus, the surface charge density on the outer surface is:
σouter=Q+q4πr22\sigma_{\text{outer}} = \frac{Q + q}{4\pi r_2^2}
B. Electric Field at a Point x>r2x > r_2
Now, let’s consider the electric field at a point outside the spherical conducting shell, i.e., at a distance xx from the center of the shell such that x>r2x > r_2.
• Since the shell is spherical and conducting, the system behaves like a point charge Qtotal=Q+qQ_{\text{total}} = Q + q located at the center of the shell.
• By Gauss's Law, the electric field outside the spherical shell (at x>r2x > r_2) is the same as if the entire charge Q+qQ + q were concentrated at the center of the shell.
Gauss's Law states:
∮E⃗⋅dA⃗=Qencϵ0\oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\epsilon_0}
For a spherical surface of radius x>r2x > r_2, the electric flux is:
E(4πx2)=Qencϵ0E(4\pi x^2) = \frac{Q_{\text{enc}}}{\epsilon_0}
where EE is the electric field at distance xx and Qenc=Q+qQ_{\text{enc}} = Q + q is the total charge enclosed by the Gaussian surface.
Solving for EE:
E=Q+q4πϵ0x2E = \frac{Q + q}{4\pi \epsilon_0 x^2}
Thus, the electric field at a point x>r2x > r_2 from the center of the spherical shell is:
E=Q+q4πϵ0x2E = \frac{Q + q}{4\pi \epsilon_0 x^2}
Summary:
• (i) The surface charge density on the inner surface of the shell is σinner=−q4πr12\sigma_{\text{inner}} = \frac{-q}{4\pi r_1^2}.
• (ii) The surface charge density on the outer surface of the shell is σouter=Q+q4πr22\sigma_{\text{outer}} = \frac{Q + q}{4\pi r_2^2}.
• (iii) The electric field at a point x>r2x > r_2 is E=Q+q4πϵ0x2E = \frac{Q + q}{4\pi \epsilon_0 x^2}.
