Two types of Potential energy for System of Charges.
For any system of charged bodies, the total electrostatic potential energy does not come from a single source. It can be understood as arising from two conceptually distinct processes:
In Electrostatics chapter, when we studied two point charges, the potential energy \( U = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{r} \) was purely interaction energy. Each charge was assumed to already exist. We only calculated the energy due to their mutual interaction.
But in a continuous conductor — like a spherical shell — the entire charge distribution must first be created. The energy required to build that distribution against electrostatic repulsion is called self-energy.
In this section, we focus strictly on calculating the self-energy of a single isolated conductor.
Consider a thin conducting spherical shell of radius \( R \), carrying total charge \( Q \). We want to determine the electrostatic self-energy of this configuration — that is, the work required to assemble this charge distribution from infinity.
When the sphere has charge \( q \), its potential is:
\[ V = \frac{q}{4\pi\varepsilon_0 R} \]To bring an additional small charge \( dq \) from infinity and deposit it on the sphere, the required work is:
\[ dU = V \, dq = \frac{q}{4\pi\varepsilon_0 R} \, dq \]Integrating from \( q = 0 \) to \( q = Q \):
\[ U = \int_0^Q \frac{q}{4\pi\varepsilon_0 R} \, dq = \frac{1}{4\pi\varepsilon_0 R} \cdot \frac{Q^2}{2} \] \[ \boxed{ U_{\text{self}} = \frac{Q^2}{8\pi\varepsilon_0 R} } \]This is the electrostatic self-energy of a charged conducting spherical shell.
The self-capacitance of an isolated sphere is:
\[ C = 4\pi\varepsilon_0 R \]Using the general capacitor energy expression:
\[ U = \frac{Q^2}{2C} \]Substituting:
\[ U = \frac{Q^2}{2(4\pi\varepsilon_0 R)} = \frac{Q^2}{8\pi\varepsilon_0 R} \]Thus, the self-energy derived from assembly matches exactly with the capacitor energy formula.
Energy in electrostatics can also be computed using field energy density:
\[ u = \frac{1}{2}\varepsilon_0 E^2 \]For a charged conducting spherical shell:
The energy is stored only in the region \( r \ge R \).
Consider a thin spherical shell of radius \( r \) and thickness \( dr \). Its volume element is:
\[ dV = 4\pi r^2 dr \]Energy in this element:
\[ dU = \frac{1}{2}\varepsilon_0 E^2 \, dV \] Substitute \( E(r) \): \[ dU = \frac{1}{2}\varepsilon_0 \left( \frac{Q}{4\pi\varepsilon_0 r^2} \right)^2 4\pi r^2 dr \] Simplifying: \[ dU = \frac{Q^2}{32\pi^2\varepsilon_0 r^4} \cdot 4\pi r^2 dr = \frac{Q^2}{8\pi\varepsilon_0} \frac{dr}{r^2} \] Now integrate from \( R \) to \( \infty \): \[ U = \frac{Q^2}{8\pi\varepsilon_0} \int_R^\infty \frac{dr}{r^2} \] \[ = \frac{Q^2}{8\pi\varepsilon_0} \left[ -\frac{1}{r} \right]_R^\infty = \frac{Q^2}{8\pi\varepsilon_0 R} \] \[ \boxed{ U = \frac{Q^2}{8\pi\varepsilon_0 R} } \]Consider a solid non-conducting sphere of radius \( R \) carrying total charge \( Q \), uniformly distributed throughout its volume.
Since the charge is distributed in the volume (not only on the surface), the electric field exists both inside and outside the sphere.
Let the sphere be built gradually by bringing thin spherical shells of charge from infinity and depositing them concentrically. It is critical that the charge dq is deposited concentrically such that the density is always uniform.
At an intermediate stage, suppose a sphere of radius \( r \) has already been assembled, containing charge:
\[ q(r) = Q \frac{r^3}{R^3} \]The charge density is uniform:
\[ \rho = \frac{Q}{\frac{4}{3}\pi R^3} \]A thin spherical layer of thickness \( dr \) contains charge:
\[ dq = \rho \cdot 4\pi r^2 dr = \frac{3Q}{R^3} r^2 dr \]The potential at radius \( r \) due to the already assembled inner charge \( q(r) \) is:
\[ V(r) = \frac{1}{4\pi\varepsilon_0} \frac{q(r)}{r} = \frac{1}{4\pi\varepsilon_0} \frac{Q r^2}{R^3} \]Work required to bring \( dq \) from infinity:
\[ dU = V(r)\, dq \] Substitute: \[ dU = \frac{1}{4\pi\varepsilon_0} \frac{Q r^2}{R^3} \cdot \frac{3Q}{R^3} r^2 dr \] \[ = \frac{3Q^2}{4\pi\varepsilon_0 R^6} r^4 dr \] Now integrate from \( r = 0 \) to \( R \): \[ U = \frac{3Q^2}{4\pi\varepsilon_0 R^6} \int_0^R r^4 dr \] \[ = \frac{3Q^2}{4\pi\varepsilon_0 R^6} \cdot \frac{R^5}{5} \] \[ \boxed{ U = \frac{3Q^2}{20\pi\varepsilon_0 R} } \]Energy density:
\[ u = \frac{1}{2}\varepsilon_0 E^2 \]Electric field:
Total energy:
\[ U = U_{\text{inside}} + U_{\text{outside}} \]The outside contribution equals the self-energy of a conducting shell:
\[ U_{\text{outside}} = \frac{Q^2}{8\pi\varepsilon_0 R} \]The inside contribution is:
\[ U_{\text{inside}} = \int_0^R \frac{1}{2}\varepsilon_0 E_{\text{inside}}^2 4\pi r^2 dr = \frac{Q^2}{40\pi\varepsilon_0 R} \] Adding: \[ U = \frac{Q^2}{8\pi\varepsilon_0 R} + \frac{Q^2}{40\pi\varepsilon_0 R} = \frac{3Q^2}{20\pi\varepsilon_0 R} \] \[ \boxed{ U = \frac{3Q^2}{20\pi\varepsilon_0 R} } \]A conducting spherical shell of radius \( R_1 \) carries a uniform charge \( q \). It is slowly expanded (without loss of charge) to a new radius \( R_2 \). Find the work performed by the electric forces during this process.
The self-energy of a charged conducting spherical shell of radius \( R \) is:
\[ U = \frac{q^2}{8\pi\varepsilon_0 R} \]Therefore,
\[ U_1 = \frac{q^2}{8\pi\varepsilon_0 R_1} \] \[ U_2 = \frac{q^2}{8\pi\varepsilon_0 R_2} \]Work done by electric forces equals the decrease in electrostatic potential energy:
\[ W_{\text{electric}} = U_1 - U_2 \] \[ = \frac{q^2}{8\pi\varepsilon_0} \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]This is expected because like charges repel. The shell naturally tends to expand under electrostatic repulsion, and the system releases stored energy as it expands.
Two thin concentric conducting spherical shells have radii \( R_1 \) and \( R_2 \) (\( R_2 > R_1 \)) and carry charges \( q_1 \) and \( q_2 \), respectively. Find:
The inner shell produces a potential at radius \( R_2 \):
\[ V_{1}(R_2) = \frac{1}{4\pi\varepsilon_0}\frac{q_1}{R_2} \]Interaction energy equals charge of outer shell times potential due to inner shell:
\[ W_{12} = q_2 V_{1}(R_2) = \frac{q_1 q_2}{4\pi\varepsilon_0 R_2} \]The system can be interpreted as two capacitors:
Now determine charges on each capacitor.
Energy stored:
\[ U_1 = \frac{q_1^2}{2C_1} \] \[ U_2 = \frac{(q_1 + q_2)^2}{2C_2} \] Substitute capacitances: \[ U_1 = \frac{q_1^2 (R_2 - R_1)} {8\pi\varepsilon_0 R_1 R_2} \] \[ U_2 = \frac{(q_1 + q_2)^2} {8\pi\varepsilon_0 R_2} \] Now add: \[ W = U_1 + U_2 \] After simplification, this reduces exactly to: \[ \boxed{ W = \frac{q_1^2}{8\pi\varepsilon_0 R_1} + \frac{q_2^2}{8\pi\varepsilon_0 R_2} + \frac{q_1 q_2}{4\pi\varepsilon_0 R_2} } \]This problem is subtle because several different physical ideas are interacting simultaneously: charge redistribution on conductors, self-energy, interaction energy, and equivalent capacitor interpretation. Let us carefully clarify the three main conceptual points.
When two conducting shells carry charges \( q_1 \) and \( q_2 \), the charges do not remain arbitrarily spread. They rearrange themselves to maintain:
For the outer shell:
Why? Because the inner shell produces field in the cavity region. To cancel field inside the metal of the outer shell, an induced charge \( -q_1 \) must appear on its inner surface.
Since total charge of outer shell is \( q_2 \), the outer surface must then carry:
\[ q_{\text{outer surface}} = q_2 - (-q_1) = q_1 + q_2 \]This redistribution is automatic due to conductor physics.
Interaction energy is:
\[ W_{12} = q_2 V_1(R_2) \]Here is the key:
Therefore, at radius \( R_2 \), the potential due to sphere 1 is:
\[ V_1(R_2) = \frac{1}{4\pi\varepsilon_0}\frac{q_1}{R_2} \]We multiply this by total charge of shell 2 (which is \( q_2 \)).
We do NOT treat shell 2 as a point charge for interaction energy, because interaction energy is computed as:
\[ \text{charge} \times \text{potential due to the other} \]The symmetry argument applies only to the field created by sphere 1.
Even though outer shell has surface charges:
Its total charge remains:
\[ q_2 \]Self-energy depends only on the total charge residing on that conductor, not on how it splits between surfaces.
For a thin conducting shell of radius \( R_2 \), the self-energy formula is:
\[ W_2 = \frac{q_2^2}{8\pi\varepsilon_0 R_2} \]This works because:
Self-energy is computed as work required to assemble total charge \( q_2 \) on that shell. Induced charge redistribution does not change the total assembly energy of that conductor.