HC Verma: Concepts of Physics - Volume 2 Electromagnetic Waves Page 339: Q3 Solution: A parallel-plate capacitor of plate-area and plate separation…

QPage 339: Q3: HC Verma: Concepts of Physics - Volume 2 Electromagnetic Waves Page 339: Q3 Solution: A parallel-plate capacitor of plate-area and plate separation…

A parallel-plate capacitor of plate-area

A

and plate separation

d

is joined to a battery of emf

\varepsilon

and internal resistance

R

t = 0

. Consider a plane surface of area

A/2

, parallel to the plates and situated symmetrically between them. Find the displacement current through this surface as a function of time.

Step-by-Step Verified Solution

**Answer:**

I_d = \frac{\epsilon}{2R} e^{-t/RC}

**Step-by-Step Explanation:** The charge on the capacitor during charging is

Q = C\epsilon(1 - e^{-t/RC})

. The electric field between the plates is

E = \frac{Q}{\epsilon_0 A}

. The flux through the surface of area

A/2

\Phi_E = E \cdot \frac{A}{2} = \frac{Q}{\epsilon_0 A} \cdot \frac{A}{2} = \frac{Q}{2\epsilon_0}

. Displacement current:

I_d = \epsilon_0 \frac{d\Phi_E}{dt} = \epsilon_0 \frac{d}{dt}\left(\frac{Q}{2\epsilon_0}\right) = \frac{1}{2}\frac{dQ}{dt}

. Since

Q = C\epsilon(1 - e^{-t/RC})

\frac{dQ}{dt} = C\epsilon \cdot \frac{1}{RC} e^{-t/RC} = \frac{\epsilon}{R} e^{-t/RC}

. Thus,

I_d = \frac{\epsilon}{2R} e^{-t/RC}

[citation:3][citation:4].

QPage 339: Q3: HC Verma: Concepts of Physics - Volume 2 Electromagnetic Waves Page 339: Q3 Solution: A parallel-plate capacitor of plate-area and plate separation…

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