Electrostatics Notes

Electrostatics

2.1 Electric Field and Potential

2.1.1 Field and Potential

• A point charge q creates an electric field \vec{E}(M) and an electric potential V(M) at each point M in space:
  • \vec{E}(M) = K \frac{q}{r^2} \vec{u}
  • V(M) = K \frac{q}{r}
  • Where r is the distance between q and M.
  • \vec{u} is a unit vector directed from q to M.

2.1.2 Superposition Principle

2.1.2.1 Point Charges

• The total charge Q = \sum{i=1}^{N} qi creates an electric field \vec{E}(M) and a potential V(M) such that:
  • \vec{E}(M) = \sum{i=1}^{N} \vec{E}i(M) \implies Ex(M) = \sum{i=1}^{N} E_{ix}(M), …
  • V(M) = \sum{i=1}^{N} Vi(M) = V1(M) + … + VN(M)
  • Where \vec{E}1 and V1 are created by the point charge q_1, etc.

Champ Created by Two Equal Charges on the Median of the Segment Separating Them

Cas 1: Opposite Charges q1 = q > 0 and q2 = -q

• Magnitudes of electric fields are equal: |\vec{E}1| = |\vec{E}2| = K \frac{|q|}{x^2 + a^2}.
• Due to symmetry, the total electric field \vec{E} = \vec{E}1 + \vec{E}2 is parallel to (yOy) and points downwards.
• |\vec{E}| = 2 |\vec{E}_1| \cos \beta with \cos \beta = \sin \alpha = \frac{a}{\sqrt{x^2 + a^2}}.
• Components:
  • E_x = 0
  • E_y = -|\vec{E}| = -K \frac{2qa}{(x^2 + a^2)^{\frac{3}{2}}}
  • \vec{E} = -K \frac{2qa}{(x^2 + a^2)^{\frac{3}{2}}} \vec{j}
• Potential: V = K \frac{q1}{r1} + K \frac{q2}{r2} = 0
• For q < 0, the electric field directions are reversed, but the magnitude calculation remains consistent.
• E_y = +|\vec{E}| = -K \frac{2qa}{(x^2 + a^2)^{\frac{3}{2}}} > 0 (since q < 0).

Cas 2: Identical Charges q1 = q2 = q

• Electric Field:
  • |\vec{E}| = 2 |\vec{E}_1| \cos \alpha, parallel to (xOx) in the positive direction.
  • E_x = |\vec{E}| = K \frac{2qx}{(x^2 + a^2)^{\frac{3}{2}}}
  • E_y = 0
  • \vec{E} = K \frac{2qx}{(x^2 + a^2)^{\frac{3}{2}}} \vec{i}
• Potential: V = 2V_1 = K \frac{2q}{\sqrt{x^2 + a^2}}
• These results hold for q < 0.

2.1.2.2 Continuous Distribution

• Each elementary charge dq is considered as a point charge, creating an elementary field d\vec{E} and potential dV.
  • d\vec{E}(M) = K \frac{dq}{r^2} \vec{u}
  • dV(M) = K \frac{dq}{r}
• In the superposition principle, sums become integrals:
  • Ex(M) = \intQ dE_x(M), …
  • V(M) = \int_Q dV(M)
  • Q = \int_Q dq
• The integral \int_Q is taken over the entire region where the charge Q is located.

Continuous Charge Distributions

• Linear Density: dq = \lambda dl
• Surface Density: dq = \sigma dS
• Volume Density: dq = \rho dV
• The choice and expression of dl, dS, and dV depend on the problem and are chosen to facilitate calculations.
• Uniform Distribution (constant density):
  • Q = \int_L \lambda dl = \lambda L
  • Q = \int_S \sigma dS = \sigma S
  • Q = \int_V \rho dV = \rho V

Examples of Charge Distributions

• Assuming a positive charge distribution, but the results are applicable to negative distributions by inverting the field directions.

Uniform Distribution on an Infinite Line (xOx)

• Due to symmetry, dEx = 0, dEz = 0, and
  • dEy = 2 |d\vec{E}1| \cos \alpha = K \frac{2dqy}{(y^2 + x^2)^{\frac{3}{2}}}
  • dq = \lambda dx
• Due to symmetry, Ex = Ez = 0, and E_y depends only on y.
• The total field \vec{E}(y) = E_y(y) \vec{j} is obtained by integrating over half the line:
  • Ey(y) = \int{0}^{\infty} K \frac{2\lambda dx y}{(y^2 + x^2)^{\frac{3}{2}}}
• Using the substitution x = y \tan \alpha, we get dx = \frac{y d\alpha}{\cos^2 \alpha} and y^2 + x^2 = \frac{y^2}{\cos^2 \alpha}.
  • Ey(y) = \int{0}^{\frac{\pi}{2}} K \frac{2\lambda \frac{y d\alpha}{\cos^2 \alpha} y}{(\frac{y^2}{\cos^2 \alpha})^{\frac{3}{2}}} = \int_{0}^{\frac{\pi}{2}} K \frac{2\lambda d\alpha \cos \alpha}{y} = K \frac{2\lambda}{y}
  • Ey(y) = \frac{\lambda}{2 \pi \epsilon0} \frac{1}{y}

Uniform Circular Distribution (Loop)

• Due to symmetry, only the component along Oz is non-zero.
  • dE_z = K \frac{2dq}{(z^2 + R^2)} \cos \alpha = K \frac{2dqz}{(z^2 + R^2)^{\frac{3}{2}}}
• Integrate over the semi-circle:
  • Ez(z) = K \frac{z}{(z^2 + R^2)^{\frac{3}{2}}} \int{demi-cercle} 2dq = K \frac{Qz}{(z^2 + R^2)^{\frac{3}{2}}}
  • Ez(z) = \frac{\lambda}{2 \epsilon0} \frac{zR}{(z^2 + R^2)^{\frac{3}{2}}}, where Q = \lambda 2\pi R
• For the potential:
  • dV = K \frac{dq}{\sqrt{z^2 + R^2}} \implies V(z) = K \frac{Q}{\sqrt{z^2 + R^2}} = \frac{\lambda}{2 \epsilon_0} \frac{R}{\sqrt{z^2 + R^2}}

Uniform Distribution on a Disc

• The ring with radius r, thickness dr, and charge dq creates a field:
  • dE_z = K \frac{dqz}{(z^2 + r^2)^{\frac{3}{2}}}
• The surface of the ring is dS = 2 \pi r dr and its charge is dq = \sigma dS
  • Ez(z) = \int{Disque} dEz = \int{0}^{R} K \frac{(\sigma 2 \pi r dr)z}{(z^2 + r^2)^{\frac{3}{2}}}
• Change variable u = z^2 + r^2 and du = 2rdr.
  • Ez(z) = K \sigma \pi z \int{z^2}^{z^2 + R^2} u^{-\frac{3}{2}} du = K \sigma \pi z [\frac{u^{-\frac{1}{2}}}{-\frac{1}{2}}]_{z^2}^{z^2 + R^2} = 2K \sigma \pi z [\frac{1}{|z|} - \frac{1}{\sqrt{z^2 + R^2}}]
  • Ez(z) = \frac{\sigma}{2 \epsilon0} [sign(z) - \frac{z}{\sqrt{z^2 + R^2}}]
• Potential:
  • dV = K \frac{dq}{\sqrt{z^2 + r^2}} = K \frac{(\sigma 2 \pi r dr)}{\sqrt{z^2 + r^2}}
  • V(z) = \int{z^2}^{z^2 + R^2} K \sigma \pi u^{-\frac{1}{2}} du = \frac{\sigma}{2 \epsilon0} [|z| - \sqrt{z^2 + R^2}]

Infinite Plane

• For an infinite plane (R \longrightarrow \infty), we have \sqrt{z^2 + R^2} \approx R and \frac{z}{\sqrt{z^2 + R^2}} \approx 0.
• The electric field becomes:
  • Ez(z) = \frac{\sigma}{2 \epsilon0} sign(z) = \begin{cases} \frac{\sigma}{2 \epsilon0}, & z > 0 \ -\frac{\sigma}{2 \epsilon0}, & z < 0 \end{cases}
• The potential becomes:
  • V(z) = \frac{\sigma}{2 \epsilon_0} [|z| - R]
• Taking the origin of potentials as arbitrary, we can remove the infinite constant -\frac{\sigma R}{2 \epsilon_0}.
  • V(z) = \frac{\sigma |z|}{2 \epsilon0} = \begin{cases} \frac{\sigma z}{2 \epsilon0}, & z \geq 0 \ -\frac{\sigma z}{2 \epsilon_0}, & z \leq 0 \end{cases}
• This is zero at z = 0.

2.2 Charges in Electric Fields and Potentials

2.2.1 Force and Potential Energy of a Point Charge

• Considering a point charge qi creating an electric field \vec{E}i(M) = K \frac{qi}{r^2} \vec{u} and a potential Vi(M) = K \frac{q_i}{r} at each point M in space.
• If another charge q is placed at M, it will be subjected to a force:
  • \vec{F}i(q) = K \frac{qqi}{r^2} \vec{u} = q \vec{E}_i(M)
• It will have potential energy:
  • E{ip}(q) = K \frac{qqi}{r} = qV(M)
• These relationships remain valid for a non-point charge Q = \sumi qi, since \vec{E}(M) = \sumi \vec{E}i(M) and \vec{F}(q) = \sumi \vec{F}i(q) = \sumi q \vec{E}i(M) = q \vec{E}(M).
• Similarly for E_p.

Force and Energy

• If a point charge q is placed at a point M with an electric field \vec{E}(M) and potential V(M), it will experience:
  • Force: \vec{F}(q) = q \vec{E}(M)
  • Potential Energy: E_p(q) = qV(M)

2.2.2 Relationship Between Field and Potential

• The force \vec{F}i(q) applied by a point charge qi to a point charge q derives from a potential:
  • \vec{F}i(q) \cdot d\vec{l} = -dE{pi}(q) \iff \vec{F}i(q) = -\nabla E{pi}(q)
• Summing over i, we get the same for the resulting force \vec{F}(q).
  • \vec{F}(q) \cdot d\vec{l} = -dEp(q) \iff \vec{F}(q) = -\nabla Ep(q)

Field and Potential

• Dividing by q, the electric field \vec{E} derives from the potential V:
  • \vec{E}(M) \cdot d\vec{l} = -dV(M) \iff \vec{E}(M) = -\nabla V(M)
  • dV(M) = V(N) - V(M), d\vec{l} = \vec{MN}
• If the potential is known, the electric field can be determined:
  • Ex(M) = -\frac{\partial V(M)}{\partial x}, Ey(M) = -\frac{\partial V(M)}{\partial y}, E_z(M) = -\frac{\partial V(M)}{\partial z}
  • E{\rho}(M) = -\frac{\partial V(M)}{\partial \rho}, E{\phi}(M) = -\frac{1}{\rho} \frac{\partial V(M)}{\partial \phi}, E_z(M) = -\frac{\partial V(M)}{\partial z}
  • Er(M) = -\frac{\partial V(M)}{\partial r}, E{\theta}(M) = -\frac{1}{r} \frac{\partial V(M)}{\partial \theta}, E_{\phi}(M) = -\frac{1}{r \sin \theta} \frac{\partial V(M)}{\partial \phi}
• Conversely, if the field is known, the potential can be determined (up to an arbitrary constant):
  • dV = -\vec{E}(M) \cdot d\vec{l} \iff V(M) = -\int \vec{E}(M) \cdot d\vec{l} \iff V(B) - V(A) = -\int_{A}^{B} E \cdot d\vec{l}
• d\vec{l} = dx \vec{i} + dy \vec{j} + dz \vec{k} = dr \vec{ur} + \rho d\phi \vec{u{\phi}} + dz \vec{k} = dr \vec{ur} + r d\theta \vec{u{\theta}} + r \sin \theta d\phi \vec{u_{\phi}}

Remarks

• The integral \int_{A}^{B} E \cdot d\vec{l} is the circulation of the vector \vec{E} along a curve from A to B.
• The circulation of the electric field is independent of the path.
• In this course, the integral will simplify:
  • If Ey = Ez = 0, then V(B) - V(A) = -\int{xA}^{xB} Ex dx
  • If E{\theta} = E{\phi} = 0, then V(B) - V(A) = -\int{rA}^{rB} Er dr

2.2.3 Internal Energy of a System of Charges

• To assemble charges qi without kinetic energy, an operator must bring them one by one from infinity at a very slow speed, applying a force equal and opposite to the electric force \vec{F}{op i} = -\vec{F}_{elec i}.
• There are three equivalent definitions of internal energy U:
  • The work done by an operator to assemble the charges one by one from a state of zero internal energy (infinity):
    • U = \sumi W{\infty \rightarrow ri}(\vec{F}{op i})
  • The work of electrostatic forces when moving the charges to positions where the potential of each charge is zero:
    • \vec{F}{op i} = -\vec{F}{elec i} \implies U = \sumi W{ri \rightarrow \infty}(\vec{F}{elec i})
  • The sum of the variations of the potential energies \Delta Ep(qi) = Ep(ri) - E_p(\infty) of each charge during assembly:
    • W{\infty \rightarrow ri}(\vec{F}{elec i}) = \Delta Ep(qi) \implies U = \sumi \Delta Ep(qi)

Internal Energy Calculations

• Internal energy of two point charges: U{12} = K \frac{q1 q2}{r} = Ep(q1) = Ep(q2) = U{21}
• Internal energy of three point charges: U{123} = U{12} + U{13} + U{23} = \frac{1}{2}(U{12} + U{21} + U{13} + U{31} + U{23} + U{32})
• Internal energy of N point charges: U = \sum_{i0, the dipole has a:
  • Potential Energy: E_p = -\vec{p} \cdot \vec{E} = -|\vec{p}| |\vec{E}| \cos(\theta)
  • Torque: \vec{\tau} = \vec{p} \land \vec{E}, |\vec{\tau}| = |\vec{p}| |\vec{E}| |\sin(\theta)|

Derivation

• Ep = EP(q) + E_p(-q) = q(V(M) - V(M')) = q(-\vec{E} \cdot \vec{M'M}) = -\vec{p} \cdot \vec{E}
• \vec{\tau} = \vec{OM'} \land \vec{F} - \vec{OM} \land \vec{F} = \vec{M'M} \land \vec{F} = \vec{M'M} \land q\vec{E} = \vec{p} \land \vec{E}

Remarks

• \vec{E} is created by other charges and should not be confused with the field created by the dipole.
• The stable equilibrium position corresponds to minimal E_p (\cos(\theta) = 1). Therefore, \vec{p} is parallel and in the same direction as \vec{E}.
• The unstable equilibrium position corresponds to maximal E_p (\cos(\theta) = -1). Therefore, \vec{p} is parallel and in the opposite direction to \vec{E}.
• In both equilibrium positions, \vec{\tau} = 0 because \vec{p} \parallel \vec{E}.

2.5 Gauss's Theorem

• Gauss's theorem allows quick calculation of the electric field created by symmetric charge distributions.

2.5.1 Flux of the Electric Field

• Subdivide a surface S into elementary surfaces dS.
• For each elementary surface, define a normal vector d\vec{S} such that |d\vec{S}| = dS > 0.
• The elementary flux of the electric field \vec{E} through dS is defined as:
  • d\Phi = \vec{E} \cdot d\vec{S} = |\vec{E}| dS \cos \theta
• The flux of the field through S = \int_S dS is
  • \Phi = \intS d\Phi = \intS \vec{E} \cdot d\vec{S}

Remarks

• If |\vec{E}| and \theta are constant on S, then \Phi = |\vec{E}| \cos \theta \int_S dS = |\vec{E}| S \cos \theta.
• If \vec{E} \parallel \pm d\vec{S}, we have:
  • \theta = 0 \implies \Phi = |\vec{E}| S \implies d\Phi = EdS, \Phi = ES
  • \theta = \pi \implies \Phi = -|\vec{E}| S \implies d\Phi = -EdS, \Phi = -ES
• By convention, if S is closed, all d\vec{S} must point outward from S.
• For a spherical surface element, dS = r^2 d\Omega, where r is the distance between dS and the sphere's center and d\Omega is the solid angle.
• In spherical coordinates, d\Omega = \sin \theta d\theta d\phi.
• For the entire sphere, \Omega = \intS d\Omega = \int{0}^{\pi} \sin \theta d\theta \int{0}^{2\pi} d\phi = 4\pi and S = \intS dS = \int_S R^2 d\Omega = 4\pi R^2.

Theorem

• Consider a charge Q distributed in space. Choose any closed geometric surface SG. Gauss's theorem states that the flux of the electric field through SG equals the charge Q{int} inside SG divided by \epsilon_0:
  • \Phi = \int{SG} \vec{E} \cdot d\vec{S} = \frac{Q{int}}{\epsilon0}

Figures

• The figures demonstrate the theorem for a single point charge, where |\vec{E}| = \frac{1}{4 \pi \epsilon_0} \frac{q}{r^2}, dS = r^2 d\Omega, and \Omega = 4\pi.

Remarks

• S_G is chosen closed but arbitrarily shaped to simplify calculations.
• If the medium is not vacuum, replace \epsilon_0 with \epsilon.

Important Special Cases

Infinite Plane

• Charge is uniformly distributed (constant density \sigma) on an infinite plane coinciding with xOy.
• SG is a cylinder with base S and lateral surface S1. The charged plane is a plane of symmetry for S_G. Assume \sigma > 0.

Symmetry

• Due to symmetry, \vec{E} = E \vec{k} with E constant on S (E depends only on z).
• Since d\vec{S} = dS \vec{k} \parallel \vec{E}, we find \Phi = \intS \vec{E} \cdot d\vec{S} + \intS (-\vec{E}) \cdot (-d\vec{S}) + \int{S1} \vec{E} \cdot d\vec{S} = 2ES + 0
• Gauss's theorem gives 2ES = \frac{Q{int}}{\epsilon0} with Q{int} = \sigma S. Therefore, E = \frac{\sigma}{2 \epsilon0} and
  • \vec{E} = \begin{cases} \frac{\sigma}{2 \epsilon0} \vec{k}, & z > 0 \ -\frac{\sigma}{2 \epsilon0} \vec{k}, & z < 0 \end{cases}

Important

• This formula is valid for negative density \sigma. It must be learned.

Sphere

• The algebraic charge Q is uniformly distributed on (or inside) a sphere S of radius R. Choose S_G = 4\pi r^2 spherical with radius r:
  • r < R for calculating the field inside the charged sphere.
  • r > R for calculating the field outside the charged sphere.

Symmetry Consideration

• Due to symmetry, \vec{E} = E \vec{ur} with E constant on SG. Since d\vec{S} = dS \vec{ur} \parallel \vec{E}, then \Phi = ESG = E 4\pi r^2 = \frac{Q{int}}{\epsilon0}.
  • \vec{E} = \frac{1}{4 \pi \epsilon0} \frac{Q{int}}{r^2} \vec{u_r}

Determine Q_{int}:

Infinite Height Cylinder

• The charge Q is uniformly distributed on the surface or inside an infinite cylinder with radius R. Choose SG = S1 + S2 + S3 a cylinder with radius r and height L:
  • S1: lateral surface, S2 = S_3: bases

Symmetry Reasoning

• Due to symmetry, \vec{E} \cdot d\vec{S2} = \vec{E} \cdot d\vec{S3} = 0, \vec{E} = E \vec{ur} with E constant on S1. Since d\vec{S1} = dS \vec{ur} \parallel \vec{E}, then \Phi = ES1 = E 2 \pi r L = \frac{Q{int}}{\epsilon_0}$
  • \vec{E} = \frac{1}{2 \pi \epsilon0} \frac{Q{int}}{rL} \vec{u_r}

Infinite Wire

• The charge Q is uniformly distributed on an infinitely long line. We proceed as for the exterior of a cylinder with Q_{int} = \lambda L.
  • \vec{E} = \frac{\lambda}{2 \pi \epsilon0} \frac{1}{r} \vec{ur}

Important Notes

• For surface distributions (sphere and cylinder), the field is zero inside. This is generally valid for any surface distribution.
• Outside the charged sphere (surface or volume), the field expression is the same as for a point charge Q.
• These remarks, along with the field created by an infinite plane, should be memorized as they are frequently used.