Question:
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.
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.
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.
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Trace the wires: all three capacitors turn out to be in parallel between A and B. To keep the same potential difference (same charge/capacitance), the capacitance of each capacitor must be unchanged.
Updated On: Nov 11, 2026
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The Correct Option is C
Solution and Explanation
Circuit: A is joined to D, and D is joined to E by the top wire, so A, D, E are at one potential. C is joined by the lower wire to the right end of C3, which goes to B, so C and B are at one potential. Thus C1 (D to C), C2 (C to E) and C3 (E to B) are each connected between the same two nodes: all three are in parallel.
The potential difference is kept the same, so the capacitance of each capacitor must not change (the equivalent capacitance of the combination stays the same).
Initially \(C = \frac{\varepsilon_0A}{d}\). After introducing a slab of thickness \(t = 3\ \mathrm{mm}\) with the gap increased to \(d + 2.4\ \mathrm{mm}\): \(C' = \frac{\varepsilon_0A}{(d + 2.4) - t + \frac{t}{K}}\).
Equating: \(d = d + 2.4 - 3 + \frac{3}{K} \Rightarrow \frac{3}{K} = 0.6 \Rightarrow K = 5\)
Answer: (3) 5
The potential difference is kept the same, so the capacitance of each capacitor must not change (the equivalent capacitance of the combination stays the same).
Initially \(C = \frac{\varepsilon_0A}{d}\). After introducing a slab of thickness \(t = 3\ \mathrm{mm}\) with the gap increased to \(d + 2.4\ \mathrm{mm}\): \(C' = \frac{\varepsilon_0A}{(d + 2.4) - t + \frac{t}{K}}\).
Equating: \(d = d + 2.4 - 3 + \frac{3}{K} \Rightarrow \frac{3}{K} = 0.6 \Rightarrow K = 5\)
Answer: (3) 5
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