A parallel plate capacitor having plate area ‘A’ and plate separation ‘d’ is joined to a battery of emf ‘V’ and internal resistance ‘R’ at t=0 . Consider a plane surface of area $A / 3,$ parallel to the plates and situated symmetrically between them. Find the displacement current through this surface as a function of time. [ The charge on the capacitor at time ‘t’ is given by $q = C V [1 - e^{\frac{t}{τ}}] w h e r e τ = C R$ ]

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$i_{d} = \frac{V}{2 R} e^{\frac{- t d}{\in_{0} A R}}$

$i_{d} = \frac{V}{3 R} e^{\frac{- t d}{\in_{0} A R}}$

$i_{d} = \frac{2 V}{3 R} e^{\frac{- 2 t d}{\in_{0} A R}}$

$i_{d} = \frac{3 V}{2 R} e^{\frac{- 3 t d}{\in_{0} A R}}$

answer is B.

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Detailed Solution

$S u r f a c e c h a r g e d e n s i t y, σ = \frac{q}{A} = \frac{C V}{A} [1 - e^{\frac{- t}{τ}}] E l e c t r i c f i e l d b / w t h e p l a t e s o f c a p a c i t o r, E = \frac{σ}{\in_{0}} = \frac{C V}{\in_{0} A} (1 - e^{\frac{- t}{τ}}) E l e c t r i c f l u x f r o m t h e g i v e n a r e a, ϕ_{E} = \frac{E A}{3} = \frac{C V}{3 \in_{0}} [1 - e^{\frac{- t}{τ}}] D i s p l a c e m e n t c u r r e n t, i_{d} = \in_{0} \frac{d ϕ_{E}}{d t} = \in_{0} \frac{d}{d t} [\frac{C V}{3 \in_{0}} (1 - e^{\frac{- t}{τ}})] = \frac{C V}{3 τ} e^{\frac{- t}{τ}} \Rightarrow i_{d} = \frac{V}{3 R} e^{\frac{- t}{C R}} \Rightarrow i_{d} = \frac{V}{3 R} e^{\frac{- t d}{\in_{0} A R}}$