Electrostatics Question 34
Question: The two metallic plates of radius $ r $ are placed at a distance $ d $ apart and its capacity is $ C $ . If a plate of radius $ r/2 $ and thickness $ d $ of dielectric constant 6 is placed between the plates of the condenser, then its capacity will be
Options:
A) $ 7C/2 $
B) $ 3C/7 $
C) $ 7C/3 $
D) $ 9C/4 $
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Answer:
Correct Answer: D
Solution:
Area of the given metallic plate A = pr2 Area of the dielectric plate $ A’=\pi {{( \frac{r}{2} )}^{2}}=\frac{A}{4} $
Uncovered area of the metallic plates $ A’’=A-A’ $
$ =A-\frac{A}{4}=\frac{3A}{4} $
The given situation is equivalent to a parallel combination of two capacitor.
One capacitor (C’) is filled with a dielectric medium (K = 6) having area $ \frac{A}{4} $ while the other capacitor (C’’) is air filled having area $ \frac{3A}{4} $
Hence $ C _{eq}=C’+C’’=\frac{K{\varepsilon _{0}}(A/4)}{d}+\frac{{\varepsilon _{0}}(3A/4)}{d} $
$ =\frac{{\varepsilon _{0}}A}{d}( \frac{K}{4}+\frac{3}{4} ) $
$ =\frac{{\varepsilon _{0}}A}{d}( \frac{6}{4}+\frac{3}{4} )=\frac{9}{4}C $
$ ( \because C=\frac{{\varepsilon _{0}}A}{d} ) $
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