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Question
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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Solution
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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