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"In theory, theory and practice are the same. In practice, they are not." ..."Albert Einstein (1879~1955, Mathematical physicist, Nobel Prize 1921-Physics)"

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 Author Topic: Relation between displacement current and electric field with R-C circuit  (Read 18673 times) 0 Members and 1 Guest are viewing this topic. Click to toggle author information(expand message area).
Fu-Kwun Hwang
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 « Embed this message on: February 15, 2011, 11:20:41 pm » posted from:Taipei,T\'ai-pei,Taiwan

The electric displacement field is defined as:
$\boldsymbol{D} = \varepsilon_0 \boldsymbol{E} + \boldsymbol{P}\ =\varepsilon \boldsymbol{E}$.

where:
ε0 is the permittivity of free space
E is the electric field intensity
P is the polarization of the medium

The folliwing use RC-circuit to show relation between displacement current and electric field.

To make it easier, the value of ε0 is assigned as 0.5

Un-check the off check box to connect the DC voltage source to the RC circuit (ON).
k is the dielectric constant $\varepsilon/\varepsilon_0$.
You can adjust voltage source Vs, resistance R, capacitance C.

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"In theory, theory and practice are the same. In practice, they are not." ..."Albert Einstein (1879~1955, Mathematical physicist, Nobel Prize 1921-Physics)"