Electrostatic energy (or potential energy, U) is the energy stored in a system of charges. It is equal to the work done (W) to assemble these charges from infinity to their final positions.
The work done to bring a set of charges q1, q2, ..., qn into position is:
Formula: U = (1/2) ∑i=1n qiV(ri)
Where V(ri) is the potential at the location of charge qi due to *all other* charges.
For a continuous charge distribution (density ρ), the sum becomes an integral:
Formula (Continuous): U = (1/2) ∫ ρ(r)V(r) dτ
This energy can also be expressed as being stored in the electric field itself, with an energy density (u) given by:
Energy Density: u = (1/2)ε0E2
The total energy is the integral of this density over all space: U = ∫all space (1/2)ε0E2 dτ
We can find the energy of a uniformly charged solid sphere (charge Q, radius R) by integrating the energy density.
Integrating u = (1/2)ε0E2 over all space (from r=0 to ∞) gives the total energy:
Total Energy (Solid Sphere): U = 3Q2 / (20πε0R)
For a hollow conducting sphere (all charge Q on surface), E=0 inside. Integrating the field density from R to ∞ gives:
Total Energy (Hollow Sphere): U = Q2 / (8πε0R)
In electrostatics, all charge on a conductor resides on its surface. The electric field just outside the surface is E = σ / ε0 (perpendicular to the surface), and the field inside is E = 0.
The charge on the surface experiences an outward-directed force. This force is due to the field created by *all other charges* on the conductor (excluding the small patch of charge itself).
The field at the surface is an average: Eavg = (Eout + Ein) / 2 = (σ/ε0 + 0) / 2 = σ / (2ε0).
The force (dF) on a small area (dA) with charge (dq = σdA) is:
dF = (dq) * Eavg = (σdA) * (σ / 2ε0) = (σ2 / 2ε0) dA
This gives the electrostatic pressure (Force per unit area) pushing outward on the conductor:
Force per unit area (Pressure): P = dF/dA = σ2 / (2ε0) = (1/2)ε0E2
A capacitor is a device that stores electrostatic energy. It consists of two conductors (plates). When a voltage (V) is applied, one plate gets charge +Q and the other gets -Q.
Capacitance (C) is the ratio of the charge (Q) stored on one plate to the potential difference (V) between the plates.
Definition: C = Q / V
The unit of capacitance is the Farad (F), where 1 Farad = 1 Coulomb / Volt.
The capacitance of a single conductor (A) is small. If we bring an uncharged conductor (B) near it, V decreases and C increases. If we then ground conductor B, the potential of A drops significantly, and its capacitance C = Q/V increases dramatically. This is the principle: two conductors, one charged and one earthed, separated by an insulator.
Even a single conductor has capacitance. It is the ratio of its charge Q to its potential V (measured relative to V=0 at infinity).
Example: Isolated Sphere (Radius R):
The potential on its surface is V = Q / (4πε0R).
Therefore, C = Q / V = Q / (Q / (4πε0R))
Capacitance (Isolated Sphere): C = 4πε0R
This is a powerful problem-solving technique in electrostatics. It allows us to solve for the potential in a region with simple boundary conditions (like a grounded conducting plane) by replacing the boundary with one or more "image charges" outside the region of interest.
This works because of the Uniqueness Theorem: if we can find a configuration of image charges that satisfies the boundary conditions (e.g., V=0 on the conductor), then the potential in the original region is solved correctly.
Problem: A point charge +q is at a distance 'd' from an infinite, grounded (V=0) conducting plane.
Solution: We remove the plane. We place an "image" charge -q at a distance 'd' *behind* the plane (at z = -d, if +q is at z = +d).
The potential in the region z > 0 is now the superposition of V+q and V-q. This configuration correctly creates V=0 on the z=0 plane, so it is the unique solution for the region z > 0.
Problem: A point charge +q is at a distance 'a' from the center of a grounded (V=0) conducting sphere of radius R (where a > R).
Solution: We replace the sphere with two image charges placed *inside* the original sphere's volume:
This arrangement of +q, q', and b produces V=0 on the surface r=R, giving the correct potential *outside* the sphere.
A dielectric is an electrical insulator. When placed in an external electric field (E0), it does not conduct, but the molecules within it become polarized.
The external field E0 causes the positive and negative charges in the dielectric's atoms/molecules to separate slightly. These tiny dipoles create an *internal* electric field (Einduced) that opposes the external field.
The net electric field (E) inside the dielectric is the vector sum: E = E0 + Einduced. The magnitude is E < E0.
Polarization (P) is a vector field defined as the net dipole moment per unit volume. It measures the degree of polarization of the material.
This polarization is not due to free-moving charges, but to "bound" charges. This creates an effective charge density:
For most materials (linear dielectrics), the polarization P is directly proportional to the *net* electric field E inside the material:
Formula: P = ε0χeE
χe (Chi) is the electric susceptibility, a dimensionless quantity that measures how easily a material polarizes.
The Dielectric Constant (κ), also called relative permittivity (εr), is defined as:
Formula: κ = 1 + χe
It relates the permittivity of the material (ε) to the permittivity of free space (ε0): ε = κε0.
In materials, the total electric field E is caused by *both* free charges (ρfree) and bound charges (ρb). This can be complicated.
We define a new auxiliary vector field, the Electric Displacement (D), which depends *only* on the free charges.
Definition: D = ε0E + P
We can write D in terms of E using the susceptibility:
D = ε0E + (ε0χeE) = ε0(1 + χe)E
Since κ = 1 + χe, we get the simple relation:
Formula: D = κε0E = εE
The main advantage of D is that it obeys a simple form of Gauss's Law that only involves the *free* charge enclosed.
Gauss's Law for D: ∮ D ⋅ dA = Qfree, enclosed
In differential form: ∇ ⋅ D = ρfree
When the space between the plates of a capacitor is filled with a dielectric material (constant κ), the capacitance is increased.
For a given charge Q on the plates, the free charge is unchanged. The dielectric polarizes, creating an opposing field Einduced.
The new capacitance C is:
C = Q / V = Q / (V0 / κ) = κ (Q / V0) = κC0
Capacitance with Dielectric: C = κC0
The capacitance is increased by a factor of κ.