question_answer

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

A)                    \[7C/2\]

B)                                      \[3C/7\]

C)                    \[7C/3\]                          

D)            \[9C/4\]

Correct Answer: D

Solution :

           Area of the given metallic plate A = pr2 Area of the dielectric plate \[A'=\pi \,{{\left( \frac{r}{2} \right)}^{2}}=\frac{A}{4}\] 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 \[{{C}_{eq}}=C'+C''=\frac{K{{\varepsilon }_{0}}\,(A/4)}{d}+\frac{{{\varepsilon }_{0}}(3A/4)}{d}\] \[=\frac{{{\varepsilon }_{0}}A}{d}\left( \frac{K}{4}+\frac{3}{4} \right)\]\[=\frac{{{\varepsilon }_{0}}A}{d}\,\left( \frac{6}{4}+\frac{3}{4} \right)=\frac{9}{4}C\]    \[\left( \because \,C=\frac{{{\varepsilon }_{0}}A}{d} \right)\]