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Q.

The distance between plates of a parallel plate capacitor is 5d. The positively charged plate is at x=0 and negativity charged plates is at x=5d. Two slabs one of conductor and the other of a dielectric of same thickness d are inserted between the plates as shown in figure. Potential (V) versus distance x graph will be

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a

b

c

d

answer is B.

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Detailed Solution

E=−dvdxE  inside the conductor is zero.
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