Electronics and Communication Engineering - Materials and Components - Discussion
Discussion Forum : Materials and Components - Section 1 (Q.No. 16)
16.
A parallel plate capacitor has its length, width and separation doubled. It fringing effects are neglected, to keep the capacitance same, the dielectric constant must be
Answer: Option
Explanation:
. When dimensions are changed, A become four times and d becomes twice.
Since ∈0 is constant, ∈r must be halved to keep C constant.
Discussion:
7 comments Page 1 of 1.
Manju power said:
3 years ago
Area=L*W.
(L*2)*(W*2) =4.
Distance d=2.
DC will be constant...assume DC. = 1
C=eA/d
= 1*4/2= 2Farads.
(L*2)*(W*2) =4.
Distance d=2.
DC will be constant...assume DC. = 1
C=eA/d
= 1*4/2= 2Farads.
Rakhul Rajumohan said:
5 years ago
@All.
It's Very simple.
Area wants to increase 4 and the distance between them becomes 2 then only we can maintain the dielectric constant same for any alternatives!
Ex:- area=4, distance = 2 then dc= 2 times, same as area =8, distance= 4 then dc= 2.
Likewise that,
Hope you understand clearly.
It's Very simple.
Area wants to increase 4 and the distance between them becomes 2 then only we can maintain the dielectric constant same for any alternatives!
Ex:- area=4, distance = 2 then dc= 2 times, same as area =8, distance= 4 then dc= 2.
Likewise that,
Hope you understand clearly.
Banoth kalyan said:
7 years ago
Please Explain the answer in detail.
Deepak said:
8 years ago
Please explain in detail.
Priti said:
8 years ago
c=nA/d {where n is dielectric constant, A is an area of the plate which multiplied of length and width and d is separation area of the plate.
C1/C2=(k1*A1*d2)/(k2*A2*d1) {where A2=4A1 AND d2=2d1 and c1=c2
then k2=k1/2.
C1/C2=(k1*A1*d2)/(k2*A2*d1) {where A2=4A1 AND d2=2d1 and c1=c2
then k2=k1/2.
(1)
Mark D. said:
8 years ago
C=k*A/d,
C=k*(2L*2W)/2d,
C=(k/2)*(2L*2W)/2d.
C=k*(2L*2W)/2d,
C=(k/2)*(2L*2W)/2d.
Riya said:
10 years ago
Can anyone explain this answer in detail once again?
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