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What is the new capacitance of the plates, when a slab of Dielectric Constant K and thickness one-fourth of the separation of plate is inserted between the plates?
  • a)
    4KCo/K + 2
  • b)
    4KCo/1 + 2K
  • c)
    4KCo/K + 3
  • d)
    4KCo/1 + 3K
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
What is the new capacitance of the plates, when a slab of Dielectric C...
As the slab is inserted between the plates,
V = E0(d/4)/K + Eo(3d/4)
= Eod (1/4K + 3/4)
= Vo(1+3K)/4K
Now we know, C = Qo/V
= 4KCo/(1 + 3K)
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Community Answer
What is the new capacitance of the plates, when a slab of Dielectric C...
As the slab is inserted between the plates,
V = E0(d/4)/K + Eo(3d/4)
= Eod (1/4K + 3/4)
= Vo(1+3K)/4K
Now we know, C = Qo/V
= 4KCo/(1 + 3K)
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