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A Parallel-plate Capacitor of Plate-area a and Plate Separation D is Joined to a Battery of Emf ε and Internal Resistance R at T = 0. Consider a Plane Surface of Area A/2,

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प्रश्न

A parallel-plate capacitor of plate-area A and plate separation d is joined to a battery of emf ε and internal resistance R at 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.

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उत्तर

Electric field strength for a parallel plate capacitor,

`E = Q/(∈_0A)`

Electric flux linked with the area,

`phi_E = EA = Q/(∈_0A) xx A/2 = Q/(2∈_0)`

Displacement current , 

`I_d = ∈_0 (dphi_E)/dt = ∈_0 d/dt(Q/(∈_0 2))`

`I_d = 1/2((dQ)/dt)`     ....(i)

Charge on the capacitor as a function of time during charging,

`Q = εC[1 - e^(-t"/"RC)]`:

Putting this in equation (i), we get :

`I_d = 1/2 εC d/dt(1 - e^(-t"/"RC))`

`I_d = 1/2 εC(-e^(-t"/"RC)) xx (-1/(RC))`

`C = (A∈_0)/d`

⇒ `I_d = ε/(2R) xx e^(-(td)/(ε_0AR))`

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अध्याय 40: Electromagnetic Waves - Exercises [पृष्ठ ३९९]

APPEARS IN

एचसी वर्मा Concepts of Physics Volume 1 and 2 [English]
अध्याय 40 Electromagnetic Waves
Exercises | Q 3 | पृष्ठ ३९९

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