Question

# Five identical equispaced conducting plates are given charges as shown in the figure. List -1 gives the labes of the sides of the plates and List -2 gives the charges List- 1List - 2I)aP)QII)dQ)2QIII)fR)3QIV)hS)5Q/3T)-2QU)-4Q/3 When switch S1 is open and S2 is closed, match List-1 with List -2

A
I P II S III U IV R
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B
I P II T III R IV Q
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C
I P II Q III T IV Q
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D
I P II S III R IV T
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Solution

## The correct option is A I→ P II → S III → U IV → R By balancing charge on A and D, Q+q−q′+q′+2Q+q=5Q -- (1) By equating potentials on A and D, Q−q′C+3Q−q′C+−q′C=0 -- (2) Solving the equations, we get q=Q and q′=4Q3 Charge on a =q=Q Charge on d =3Q−q′=5Q3 Charge on f =−q′=−4Q3 Charge on h =2Q+q=3Q

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