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Electric Field and Electric Potential

Electric Field and Electric Potential

(Electrostatics – Physics Honours Notes)


📌 1. Electric Field (𝐄⃗ )

Electric field represents the influence of a charge on the surrounding space.

Definition

Electric field at a point is the force experienced per unit positive test charge.E=Fq\vec{E} = \frac{\vec{F}}{q}E=qF​

Electric Field Lines

  • Represent direction of electric field.
  • Originates from positive and ends on negative charges.
  • Never intersect.
  • Density of lines = strength of electric field.

Electric Flux

Flow of electric field lines through a surface.ϕE=EA=EAcosθ\phi_E = \vec{E} \cdot \vec{A} = EA \cos\thetaϕE​=E⋅A=EAcosθ

Measured in N·m²/C


📌 2. Gauss’s Law

Total electric flux through a closed surface is equal to charge enclosed divided by ε₀:EdA=Qenclosedϵ0\oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enclosed}}}{\epsilon_0}∮E⋅dA=ϵ0​Qenclosed​​

Applications of Gauss’s Law


⭐ A) Spherical Symmetry — Charged Sphere

  1. Point charge or uniformly charged sphere (outside sphere)

E=14πϵ0Qr2E = \frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}E=4πϵ0​1​r2Q​

  1. Inside a uniformly charged sphere

E=Qr4πϵ0R3E = \frac{Qr}{4\pi\epsilon_0 R^3}E=4πϵ0​R3Qr​

(∝ r ⇒ zero at center)


⭐ B) Cylindrical Symmetry — Infinite Line of Charge

E=λ2πϵ0rE = \frac{\lambda}{2\pi\epsilon_0 r}E=2πϵ0​rλ​


⭐ C) Planar Symmetry — Infinite Plane Sheet

E=σ2ϵ0E = \frac{\sigma}{2\epsilon_0}E=2ϵ0​σ​

(Independent of distance!)


📌 3. Conservative Nature of Electrostatic Field

  • Work done in a closed path = 0

Edl=0\oint \vec{E} \cdot d\vec{l} = 0∮E⋅dl=0

  • Electric field = negative gradient of potential

E=V\vec{E} = -\nabla VE=−∇V


📌 4. Electric Potential (V)

Work done by external force in bringing a unit positive charge from infinity to a point.V=WqV = \frac{W}{q}V=qW​

Potential due to a Point Charge

V=14πϵ0QrV = \frac{1}{4\pi\epsilon_0} \frac{Q}{r}V=4πϵ0​1​rQ​


Electric Potential and Field of a Dipole

Electric dipole: two equal & opposite charges separated by distance 2a
Dipole moment:p=q2a\vec{p} = q \cdot 2ap​=q⋅2a

On axial line:Vaxial=14πϵ0pr2V_{\text{axial}} = \frac{1}{4\pi\epsilon_0} \frac{p}{r^2}Vaxial​=4πϵ0​1​r2p​

On equatorial line:Vequatorial=0V_{\text{equatorial}} = 0Vequatorial​=0

Force on dipole in uniform field:F=0\vec{F} = 0F=0

Torque:τ=p×E\vec{\tau} = \vec{p} \times \vec{E}τ=p​×E


📌 5. Laplace & Poisson Equations

2V=0(Laplace in charge-free region)\nabla^2 V = 0 \quad (\text{Laplace in charge-free region})∇2V=0(Laplace in charge-free region) 2V=ρϵ0(Poisson with charge)\nabla^2 V = -\frac{\rho}{\epsilon_0} \quad (\text{Poisson with charge})∇2V=−ϵ0​ρ​(Poisson with charge)


📌 6. Uniqueness Theorem

The solution of electrostatic potential satisfying:

  1. Poisson/Laplace equation and
  2. Boundary conditions
    is unique.

📌 7. Method of Images

Used to solve problems involving conductors by replacing boundaries with imaginary charges.

Applications

1️⃣ Point charge near infinite grounded conducting plane
→ Replace by image charge of equal magnitude but opposite sign behind plane.
Helps calculate:

  • Potential
  • Force on charge
  • Surface charge density

2️⃣ Charge near grounded conducting sphere
→ Image charge placed inside sphere at a specific location
Ensures potential on surface remains zero.


📌 8. Electrostatic Energy

System of Point Charges

U=12iqiViU = \frac{1}{2}\sum_{i} q_i V_iU=21​i∑​qi​Vi​

Charged Isolated Sphere

U=12Q24πϵ0RU = \frac{1}{2} \frac{Q^2}{4\pi\epsilon_0 R}U=21​4πϵ0​RQ2​

ORU=12QVU = \frac{1}{2} QVU=21​QV


📌 9. Conductors in Electrostatic Field

  • Electric field inside a conductor = 0
  • Excess charge resides only on surface
  • Surface is an equipotential
  • Surface charge density:

σ=ϵ0E\sigma = \epsilon_0 E_\perpσ=ϵ0​E⊥​

Force on conductor surface:f=σ22ϵ0f = \frac{\sigma^2}{2\epsilon_0}f=2ϵ0​σ2​


✨ Key Highlights (Quick Revision)

ConceptKey Result
E-field from potentialE=V\vec{E} = -\nabla VE=−∇V
Gauss’s lawEdA=Qϵ0\oint \vec{E}\cdot d\vec{A} = \frac{Q}{\epsilon_0}∮E⋅dA=ϵ0​Q​
Infinite sheet fieldE=σ2ϵ0E = \frac{\sigma}{2\epsilon_0}E=2ϵ0​σ​
Inside charged sphereErE \propto rE∝r
ConductorsEinside=0E_{\text{inside}} = 0Einside​=0
Dipole torqueτ=pEsinθ\tau = pE\sin\thetaτ=pEsinθ

📘 Conclusion

This chapter explains how charges create fields and potentials in space, and how energy and forces are stored or applied through electrostatic interactions. Gauss’s law and boundary conditions provide powerful tools to solve symmetric electrostatic problems effectively.

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