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A parallel plate capacitor of a capacitance 1 pF has seperation between the plates is d. When the distance of seperation becomes 2d and wax of dielectric constant x is inserted in it the capacitance becomes 2 pF. What is the value of x?
Options
(a) 2
(b) 4
(c) 6
(d) 8
Correct Answer:
4
Explanation:
No explanation available. Be the first to write the explanation for this question by commenting below.
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Topics: Electrostatics
(146)
Subject: Physics
(2479)
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- A radioactive substance decays to 1/16th of its initial activity in 40 days.
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Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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C=Akepslan not/d
If 1pf=k/d
So relation,
2pf=4x/2d so it can com 2pf.
Bhaiya dekho salon ne dala nahi tha explanation hum likh rahe hain.
Before C =1pF
After when d is doubled and dilectric is inserted
C=2pF
So x-
Since C=(ε0.A/d)k after intertion of dielectric
2=(ε0.A/2d)k (given)
2=(C/2)k
2=k/2. (Given)
K=4