Electrostatics Question 815
Question: A parallel plate capacitor of area a plate separation d is filled with two dielectrics as shown. What is the capacitance of the arrangement?
Options:
A) $ \frac{3K{\varepsilon _{0}}A}{4d} $
B) $ \frac{4K{\varepsilon _{0}}A}{4d} $
C) $ \frac{( K+1 ){\varepsilon _{0}}A}{2d} $
D) $ \frac{K( K+3 ){\varepsilon _{0}}A}{2( K+1 )d} $
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Answer:
Correct Answer: D
Solution:
[d] $ c _{1}=\frac{( A/2 ){\varepsilon _{0}}}{d/2}=\frac{A{\varepsilon _{0}}}{d}, $
$ c _{2}=K\frac{A{\varepsilon _{0}}}{d},c _{3}=K\frac{A{\varepsilon _{0}}}{2d} $
$ \therefore c _{eq}=\frac{c _{1}\times c _{2}}{c _{1}+c _{2}}+c _{3}=\frac{( 3+K )KA{\varepsilon _{0}}}{2d( K+1 )} $ ($ \therefore C _{1} $ and $ C _{2} $ are in series and resultant of these two in parallel with $ C _{3} $ )
BETA
