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C B · d s = μ 0 I + ε 0 μ 0 d Φ E d t

and is independent of the surface S through which the current I is measured.

We can now examine this modified version of Ampère’s law to confirm that it holds independent of whether the surface S 1 or the surface S 2 in [link] is chosen. The electric field E corresponding to the flux Φ E in [link] is between the capacitor plates. Therefore, the E field and the displacement current through the surface S 1 are both zero, and [link] takes the form

C B · d s = μ 0 I .

We must now show that for surface S 2 , through which no actual current flows, the displacement current leads to the same value μ 0 I for the right side of the Ampère’s law equation. For surface S 2 , the equation becomes

C B · d s = μ 0 d d t [ ε 0 Surface S 2 E · d A ] .

Gauss’s law for electric charge requires a closed surface and cannot ordinarily be applied to a surface like S 1 alone or S 2 alone. But the two surfaces S 1 and S 2 form a closed surface in [link] and can be used in Gauss’s law. Because the electric field is zero on S 1 , the flux contribution through S 1 is zero. This gives us

Surface S 1 + S 2 E · d A = Surface S 1 E · d A + Surface S 2 E · d A = 0 + Surface S 2 E · d A = Surface S 2 E · d A .

Therefore, we can replace the integral over S 2 in [link] with the closed Gaussian surface S 1 + S 2 and apply Gauss’s law to obtain

S 1 B · d s = μ 0 d Q in d t = μ 0 I .

Thus, the modified Ampère’s law equation is the same using surface S 2 , where the right-hand side results from the displacement current, as it is for the surface S 1 , where the contribution comes from the actual flow of electric charge.

Displacement current in a charging capacitor

A parallel-plate capacitor with capacitance C whose plates have area A and separation distance d is connected to a resistor R and a battery of voltage V . The current starts to flow at t = 0 . (a) Find the displacement current between the capacitor plates at time t . (b) From the properties of the capacitor, find the corresponding real current I = d Q d t , and compare the answer to the expected current in the wires of the corresponding RC circuit.

Strategy

We can use the equations from the analysis of an RC circuit ( Alternating-Current Circuits ) plus Maxwell’s version of Ampère’s law.

Solution

  1. The voltage between the plates at time t is given by
    V C = 1 C Q ( t ) = V 0 ( 1 e t / R C ) .

    Let the z -axis point from the positive plate to the negative plate. Then the z -component of the electric field between the plates as a function of time t is
    E z ( t ) = V 0 d ( 1 e t / R C ) .

    Therefore, the z-component of the displacement current I d between the plates is
    I d ( t ) = ε 0 A E z ( t ) t = ε 0 A V 0 d × 1 R C e t / R C = V 0 R e t / R C ,

    where we have used C = ε 0 A d for the capacitance.
  2. From the expression for V C , the charge on the capacitor is
    Q ( t ) = C V C = C V 0 ( 1 e t / R C ) .

    The current into the capacitor after the circuit is closed, is therefore
    I = d Q d t = V 0 R e t / R C .

    This current is the same as I d found in (a).
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Maxwell’s equations

With the correction for the displacement current, Maxwell’s equations take the form

E · d A = Q in ε 0 ( Gauss’s law )
B · d A = 0 ( Gauss’s law for magnetism )
E · d s = d Φ m d t ( Faraday’s law )
B · d s = μ 0 I + ε 0 μ 0 d Φ E d t ( Ampère-Maxwell law ) .

Once the fields have been calculated using these four equations, the Lorentz force equation

Practice Key Terms 2

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Source:  OpenStax, University physics volume 2. OpenStax CNX. Oct 06, 2016 Download for free at http://cnx.org/content/col12074/1.3
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