Question

Four identical metallic plates (initially uncharged), each having area of cross section A are each separated by a distance of d as shown in the figure above. Plate 2 is given a charge of Q Plates 1 and 4 are both earthed via a very thin conducting wire. Find the potential difference between plates 2 and 3.

Answer 1

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Answer 2

Answer:

I have redrawn the diagram and labelled the plates.

Let charge q is induced on plate A. Then an equal and opposite charge induces on left surface of plate B i.e −q . Now since plate B is isolated from other two plates so net charge on it must remain conserved therefore charge induced on right surface of B is Q−q . Hence the charge induced on left side of plate C is −(Q−q) . Let area of each plate be A0

Now potential difference between plate A and plate B is,

VAB= $\frac{q⋅aA}{0⋅ϵ0 }$

Now potential difference between plate B and plate C is,

VBC= $\frac{(Q−q)⋅b}{A0⋅ϵ0 }$

Now since plate A and plate B are connected via wire therefore they must be at same potential, i.e VAC=0

⟹VA−VC=0

⟹(VA−VB)+(VB−VC)=0

⟹VAB+VBC=0

⟹ $\frac{q⋅a}{A0ϵ0}+ \frac{(Q−q)⋅b}{A0ϵ0}=0$

⟹ $\boxed{- Q = \frac{q⋅b}{a+b}}$

Now charge on inner surface of rightmost plate (i.e plate C) is −(Q−q) .

Hence the answer is$\boxed{ - \frac{Q</p><p>a}{a+b.} }$