Conductors, Insulators, and Electrostatic Induction Notes

Conductors and Insulators

  • Conductors contain non-localized charges that can move under the influence of an electric field EE. E≠0E \neq 0
  • Insulators do not contain free charges that can move.

CEES (Charged Mobile Entities)

  • A conductor that, under certain conditions, exhibits charge displacement properties.
  • In a static situation, the electric field inside the volume of a conductor is zero. E=0E = 0
  • The electric potential is identical at all points within the conductor; this volume is equipotential.
  • Charges of a capacitor redistribute on the surfaces of the conductor.
  • Surface charge density is given by: E=qϵ0E = \frac{q}{\epsilon_0}

Grounding a Conductor

  • The ground (earth) acts as a source or sink of electrons depending on the conductor's intrinsic charge state.
    • If the conductor needs to lose electrons, the earth acts as a receiver.
    • If the conductor needs to gain electrons, the earth acts as a donor.

Electric Fields

  • E(r > a): E = 0 (for a neutral conductor)
  • E(b < r < c): \oint{E \cdot dS} = \frac{q{in}}{\epsilon0}, and E⋅S=2Qϵ<em>0  ⟹  E=2Q4πϵ</em>0r2E \cdot S = \frac{2Q}{\epsilon<em>0} \implies E = \frac{2Q}{4\pi\epsilon</em>0 r^2}
  • E(r > c): E = 0 (for an isolated conductor, not connected to ground, capable of retaining an electric charge)

Cavity within a Conductor

  • A conductor with two surfaces, internal and external.
  • ΔV=a−∮E⋅dl=0??\Delta V = a - \oint{E \cdot dl} = 0 ?? because E=0E = 0, then V<em>A=V</em>BV<em>A = V</em>B

Phenomenon of Induction

  • Consider an isolated neutral conductor under the effect of an external electric field EE. A charge distribution occurs, and the total charge remains zero: Q=0Q = 0
  • Illustration of charge redistribution with positive and negative charges.
Types of Induction
  • Partial Induction
    • 'A' is a conductor with charge QA≠0Q_A \neq 0, thus an electric field EE is created.
    • 'B' is a neutral conducting object.
    • Under the effect of EE created by A, a charge redistribution occurs such that: E<em>AE<em>A, ∣Q</em>1∣=∣Q2∣|Q</em>1| = |Q_2|
  • Total Induction
    • 'A' is an insulating sphere charged with Q_A > 0
    • 'B' is a neutral conductor.

Exercise

  1. Potentials and electric fields:
  • V<em>a−V</em>0=−∫E⋅drV<em>a - V</em>0 = -\int{E \cdot dr}
  1. Calculations:
  • E(r > c): \oint{E \cdot dS} = \frac{q{in}}{\epsilon0}; E⋅S=K(Q+2Q)r2E \cdot S = \frac{K(Q + 2Q)}{r^2}; E=3KQr2E = \frac{3KQ}{r^2}
  1. Electric Potentials:
  • V<em>a−V</em>b=−∫E⋅dr=−∫KQr2=KQ[1r]ab=KQ(1b−1a)V<em>a - V</em>b = -\int{E \cdot dr} = -\int{\frac{KQ}{r^2}} = KQ \left[ \frac{1}{r} \right]_a^b = KQ \left( \frac{1}{b} - \frac{1}{a} \right)
  • V<em>B−V</em>c=−kQCcV<em>B - V</em>c = -\frac{kQC}{c}
  • V<em>b−V</em>e=−KQV<em>b - V</em>e = -KQ (since it's a conductor)
  • V<em>E−V</em>D=−∫DEE⋅drV<em>E - V</em>D = -\int_D^E{E \cdot dr}
  • V<em>E−V</em>p=−∫<em>bc−2KQr2dr=−2KQ[1r]</em>bc=−2KQ(1c−1b)=−2KQc+2KQbV<em>E - V</em>p = -\int<em>b^c{\frac{-2KQ}{r^2} dr} = -2KQ \left[ \frac{1}{r} \right]</em>b^c = -2KQ \left( \frac{1}{c} - \frac{1}{b} \right) = \frac{-2KQ}{c} + \frac{2KQ}{b}
  • V<em>b−V</em>e(r=a)=−2KQc+2KQbV<em>b - V</em>e (r = a) = -\frac{2KQ}{c} + \frac{2KQ}{b}
  • V<em>0=V</em>EV<em>0 = V</em>E (conductor)
Surface Charges
  • σb=+2Q4πb2\sigma_b = +\frac{2Q}{4 \pi b^2}
  • σc=−Q4πc2\sigma_c = -\frac{Q}{4 \pi c^2}

Homework Questions

  1. Multiple-choice questions regarding electric fields (E) in regions between spheres A, B, C, D, and E with given charge configurations.
  2. A sphere within a non-conducting sphere.
    • E<em>total=E</em>ext+Eint=0E<em>{total} = E</em>{ext} + E_{int} = 0
    • Charges: +2Q,−Q+2Q, -Q