Skip to content

prelecture_07_conductors_and_capacitance

Kurt Robert Rudolph edited this page Jun 19, 2012 · 50 revisions

Conductors and Capacitance

Overview

  • Today will extend our study of electric potential to
    • Conductors
    • Capacitance
      • We will introduce this topic
  • Last time we introduced Electric Potential
    • [\Delta V_{A \rightarrow B} = -\int\limits_{A}^{B}{ \vec E \cdot d \vec l}]
    • All example shown had fixed charge distrobutions, today we will introduce conductors without a fixed charge distrobution. * Conductors where the charges are free to move
  • After discussing non-uniform charge distrobution in a conductor, we go on to discuss capacitance
    • Specifically we discuss parallel plates
    • [C \equiv \frac{ Q}{ \Delta V}]
      • Field betwee plates [E = \frac{ Q}{ \varepsilon_0 A}]

Conductors and Equipotentials

  • [V_B - V_A = -\int\limits_{A}^B}{ \vec E \cdot d \vec l}]
  • [V_B - V_A = 0]
    • Since [\vec E = 0] everywhere within the conductor
    • This implies that the entire conductor it self is an equipotential

  • When the conductor is circularly symmetric, the field lines are perpendicular to the surface

  • When we change the shape such that it is no longer circularly symmetric the charge density redistributes such that we have varying points of charge density

Question 1

A solid elliptical conductor with a net charge [Q] is surrounded by a hollow spherical conductor with net charge [-Q]. Points [a] and [c] are on the outer surface of the inner conductor, and points [b] and [d] are on the inner surface of the outer conductor as shown.

Compare [V_{ab}], the magnitude of the potential difference between points [a] and [b], to [V_{cd}], the magnitude of the potential difference between points [c] and [d].

  • [V_{ab} = V_{cd}]
    • Since each conductor is an equipotential, the potential difference between any point on the inner conductor and any point on the outer conductor has to be the same.

Equipotential Example

  • Consider two conducting spheres which are seprerated by a large distance and have radius' [R_A] and [R_B = 4 R_A]

  • Calculating the potential at the surface of [A]

    • [V_A = -\int\limits_{\infty}^{R_A}{ \vec E_A \cdot d \vec l}]
    • [V_A = -\int\limits_{\infty}^{R_A}{ k \frac{ Q}{ r^2} d r}]
    • [V_A = k \frac{ Q}{ R_A}]
  • Repeat for sphere [B]

    • [V_B = k \frac{ Q}{ R_B}]
    • [V_B = k \frac{ Q}{ 4 R_A}]
    • [V_B = \frac{ 1}{ 4} V_A]

  • What if we connect the sphere's via a thin wire?
    • The charges will in essence become a single conductor
    • Their charges will flow until the come to the same potential.
    • [V_A = V_B]
    • [k \frac{ Q_A}{ R_A} = k \frac{ Q_B}{ R_B}]
    • [k \frac{ Q_A}{ R_A} = k \frac{ Q_B}{ 4 R_A}]
    • [\frac{ Q_A}{ R_A} = \frac{ Q_B}{ 4 R_A}]
    • [Q_A = \frac{ Q_B}{ 4}]
    • [ 4 Q_A = Q_B]
    • [ Q_A + Q_B = 2 Q]
      • [ Q_A = \frac{ 2}{ 5} Q]
      • [ Q_B = \frac{ 8}{ 5} Q]
    • Note
      • Even though sphere [B] has four times the charge of sphere [A] the charge density on sphere [A] is still larger.
      • [4 Q_A = Q_B]
      • [4 \sigma_A Area_A = \sigma_B Area_B ]
      • [4 \sigma_A 4 \pi R_A^2 = \sigma_B 4 \pi R_B^2 ]
      • [4 \sigma_A R_A^2 = \sigma_B R_B^2 ]
      • [4 \sigma_A R_A^2 = \sigma_B (4 R_A)^2 ]
      • [4 \sigma_A R_A^2 = \sigma_B 4^2 R_A^2 ]
      • [4 \sigma_A = \sigma_B 4^2]
      • [\sigma_A = \sigma_B 4]
      • [\sigma_A = 4 \sigma_B]

Charge Distribution on Conductors

  • Uncharged spherical conducting shell with a positive charge placed in the cavity.

  • The presence of the positive charge will cause the charges in the conducting shell to move so as to preserve the zero electric field in the center * QUESTION: This was an uncharged spherical shell, why is it worded this way in the prelecture??
    • Consequently both surfaces of the shell receive an induced charge.
    • [Q_{inner} = -q]
    • [Q_{outer} = q]

  • Should the charge at the center of the sphere move to the right
    • Inner surface charge distribution changes accordingly.
    • Outer surface charge distribution stays the same.

Question 2

Consider a charged solid conducting sphere of radius [a] inside an uncharged hollow conducting sphere having inner radius [b] and outer radius [c], as shown.

Which of the following graphs best describes the magnitude of the electric field as a function of distance [r] from the center of the sphere?

  • A
    • According to Gauss law the [E]-field will simply diminish as [\frac{ 1}{ r^2}] when is [r > a]. Since the outer conducting shell is uncharged it does not contribute to [E] in any way other than ensuring that [E = 0] inside the shell itself (i.e., for [b < r < c]).

Shielding in a conductor with a Cavity

  • We just explored what happens when we move a charge around the inside of a conducting spherical shell.

  • Now we will explore what happens when you do the same outside the shell.

  • Similarly the charge distribution on the outer shell will act accordingly to the position of the outer charge [q].

  • Now consider if we create a cavity within the solid conductor

  • Similarly again, the charge on the inner cavity remains unchanged by presence of the charge outside the conducting shell.
  • The inner charge field of the conducting sphere always remains zero and the charges position them selves in such a manor to act accordingly.

Capacitance

  • Let
    • [\vec E] be the electric field between the plates
    • [\Delta V] non-zero potential difference which develops as a result of this field.
  • Since
    • [E \propto Q]
  • Then
    • [\Delta V \propto Q]
  • So we define the capacitance to be
    • [C \equiv \frac{ Q}{ \Delta V}]
    • The units of capacitance are Coulomb per Volt which is a Farad
      • [ F = \frac{ C}{ V}]

  • Let
    • [A] be the surface area of the conducting plates
    • [d] be the distance between the plates
    • [L] be the edge lengths of the plates
    • [\vec E] be the electric field created between the plates
  • Given
    • The magnitude of the [E] field created separately
      • [E_{Top Plate} = \frac{ 1}{ 2} \frac{ \sigma}{ \varepsilon_0}]
      • [E_{Bottom Plate} = \frac{ 1}{ 2} \frac{ \sigma}{ epsilon_0}]
      • [E_{Top Plate} = \frac{ 1}{ 2} \frac{ Q}{ \varepsilon_0 A}]
      • [E_{Bottom Plate} = \frac{ 1}{ 2} \frac{ Q}{ epsilon_0 A}]
      • Since directions of their fields is the same, we determine that [E = \frac{ 1}{ 2} \frac{ Q}{ \varepsilon_0 A}]
    • [ d << L]
Clone this wiki locally