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NEET PHYSICSMedium

A parallel plate capacitor with cross-sectional area A and separation d has air between the plates. An insulating slab of the same area but the thickness of d/2 is inserted between the plates as shown in the figure, having a dielectric constant, K=4. The ratio of the new capacitance to its original capacitance will be:

A

2:1

B

8:5

C

6:5

D

4:1

Step-by-Step Solution

  1. Original Capacitance: With air between the plates, the capacitance is C0=ε0AdC_0 = \frac{\varepsilon_0 A}{d}.
  2. New Capacitance: When a dielectric slab of thickness tt and dielectric constant KK is inserted, the new capacitance is given by: C′=ε0A(d−t)+tKC' = \frac{\varepsilon_0 A}{(d-t) + \frac{t}{K}}
  3. Substitute Values: Here, t=d2t = \frac{d}{2} and K=4K = 4. C′=ε0A(d−d2)+d/24=ε0Ad2+d8C' = \frac{\varepsilon_0 A}{(d - \frac{d}{2}) + \frac{d/2}{4}} = \frac{\varepsilon_0 A}{\frac{d}{2} + \frac{d}{8}}
  4. Simplify: The denominator is 4d+d8=5d8\frac{4d + d}{8} = \frac{5d}{8}. C′=ε0A5d8=85(ε0Ad)=85C0C' = \frac{\varepsilon_0 A}{\frac{5d}{8}} = \frac{8}{5} \left( \frac{\varepsilon_0 A}{d} \right) = \frac{8}{5} C_0
  5. Ratio: The ratio C′C0=85\frac{C'}{C_0} = \frac{8}{5}.
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