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Q.

Two charges Q1 and Q2 are distance d apart. Two dielectrics of thickness t1 and t2 and dielectric constant k1 and k2 are introduced as shown. Find the force between the charges

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a

Q1Q24πε0d−t1+t2+k1t1+k2t22

b

zero

c

Q1Q24πε0d+k1t1+k2t22

d

Q1Q24πε0k1−1t1+k2−1t2+d2

answer is D.

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Detailed Solution

Effective distance in vacuum is given byr=t1K1+t2K2+d-t1+t2 so, F=Q1Q24πε0t1K1+t2K2+d-t1-t22 ⇒F =Q1Q24πε0K1-1t1+K2-1t2+d2
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