Capacitor Problem: Charges and Potential Differences on Metallic Plates

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SUMMARY

The discussion centers on a capacitor problem involving four metallic plates, where Plate 2 is charged with Q and Plates 1 and 4 are earthed. The charge on the right side of Plate 3 is determined to be +Q/4, while the charge on the right side of Plate 4 is -Q/4. The potential difference between Plates 1 and 2 is calculated as 3Qd/2εA. The solution involves understanding charge induction and potential equalization among the plates.

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capacitor problem ..help please!

Homework Statement


Four metallic plates are placed as shown (attachment). Plate 2 is given charge Q whereas all other plates are uncharged initially .Plate 1 and 4 are earthed.The area(A) of each plate is same .
q1) The charge appearing on the right side of plate 3 is--
a) zero b) +Q/4 c) -3Q/4 d) Q/2

q2) The charge appearing on the right side of plate 4 is--
a)zero b) -Q/4 c) -3Q/4 d) -Q/4

q3) The potential difference between plates 1 and 2 is--
a) 3Qd/2εA b)Qd/εA c)3Qd/4εA d)3Qd/εA

Homework Equations



Well I tried taking the charge on left side of plate 2 to be q and the other side of it to be Q-q...then i proceed by applying charges on all plates by induction ...but get q=Q...which is not right ...help!



The Attempt at a Solution

 

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Options of question 2 not correct and solution is that charge will flow from plate 4 to 1 due to high potential than 1. After equal potential at both plates charge +3Q/4 will be at 1 and +Q/4 will be at 4. so charge appearing on the right side of plate 3 will be +Q/4.
 

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