CBSE Class 12 · 2013 · EM WavesShort Answer

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A capacitor made of two parallel plates, each of area A and separation d, is being charged by an external ac source. Show that the displacement current inside the capacitor is the same as the conduction current charging it.

Answer: $i_{d} = ε_{0} \frac{d Φ _{E}}{d t} = \frac{d q}{d t} = i_{c}$ .

Solution

Let the instantaneous charge be q, giving surface charge density

σ = q / A

.
The uniform field between the plates is

E = \frac{σ}{ε _{0}} = \frac{q}{ε _{0} A}

The electric flux through the area between the plates is

Φ_{E} = E A = \frac{q}{ε _{0}}

The displacement current is

i_{d} = ε_{0} \frac{d Φ _{E}}{d t} = ε_{0} \frac{d}{d t} (\frac{q}{ε _{0}}) = \frac{d q}{d t} = i_{c}

Thus the displacement current between the plates equals the conduction current charging the capacitor.

i_{d} = d q / d t = i_{c}

CBSE Class 12 Board · 2013 · Subjective

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