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Capacitor question

2021 · 16 Mar · Shift 1 · Q52
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  5. /2021 · 16 Mar · Shift 1 · Q52

Capacitor question

2021 · 16 Mar · Shift 1 · Q52

JEE MainPhysicsCapacitorMCQ+4 / −1
For changing the capacitance of a given parallel plate capacitor, a dielectric material of dielectric constant K is used, which has the same area as the plates of the capacitor. The thickness of the dielectric slab is 34{3 \over 4}43​ d, where 'd' is the separation between the plates of parallel plate capacitor. The new capacitance (C') in terms of original capacitance (C0) is given by the following relation :
  1. A
    C′=3+K4KC0C' = {{3 + K} \over {4K}}{C_0}C′=4K3+K​C0​
  2. B
    C′=4+K3C0C' = {{4 + K} \over {3}}{C_0}C′=34+K​C0​
  3. C
    C′=4KK+3C0C' = {{4K} \over {K + 3}}{C_0}C′=K+34K​C0​
  4. D
    C′=43+KC0C' = {{4} \over {3 + K}}{C_0}C′=3+K4​C0​
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