JEE Main 2017 · 9 Apr · Shift 1 · Physics Q15

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PhysicsSingle correct+4 / -1Question 15 of 90

A combination of parallel plate capacitors is maintained at a certain potential difference.

When a 3 mm thick slab is introduced between all the plates, in order to maintain the same potential difference, the distance between the plates is increased by 2.4 mm. Find the dielectric constant of the slab.

Capacitors C₁ (between D and C), C₂ (between C and E) and C₃ (between E and B) in a row from terminal A to terminal B; a wire joins D to E above, and another joins C to B below.

Answer and worked solution

Official answer: C

D is joined to E, and C to B, so C₁, C₂ and C₃ all lie between the same two nodes: they are in parallel, with the same plate separation $d$ . To keep each capacitance unchanged with the slab,
$d=d+2.4-3+\frac3K\implies\frac3K=0.6,\quad K=5.$

Topics:Dielectrics Series and parallel combinations

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JEE Main 2017 · 9 Apr · Shift 1, Physics question 15