question_answer

In the figure a capacitor is filled with dielectrics. The resultant capacitance is:

A) \[\frac{2A{{\varepsilon }_{o}}}{\operatorname{d}}\left[ \frac{1}{{{k}_{1}}}+\frac{1}{{{k}_{2}}}+\frac{1}{{{k}_{3}}} \right]\]

B) \[\frac{A{{\varepsilon }_{o}}}{\operatorname{d}}\left[ \frac{1}{{{k}_{1}}}+\frac{1}{{{k}_{2}}}+\frac{1}{{{k}_{3}}} \right]\]

C)  \[\frac{2A{{\varepsilon }_{o}}}{\operatorname{d}}\left[ {{k}_{1}}+{{k}_{2}}+{{k}_{3}} \right]\]

D) \[\frac{A{{\varepsilon }_{o}}}{\operatorname{d}}\left[ \frac{{{k}_{3}}}{2}+\frac{{{k}_{1}}{{k}_{2}}}{{{k}_{1}}+{{k}_{3}}} \right]\]

Correct Answer: D

Solution :

\[\operatorname{ma} = mg sin\theta \] \[\operatorname{a} = g sin\theta \] \[\operatorname{S}=ut+\frac{1}{2}a{{t}^{2}}\] u = 0                            (starts from rest) \[\ell =0+\frac{1}{2}g\,\,sin\theta {{t}^{2}}\] \[\operatorname{t}=\sqrt{\frac{2\ell }{g\,sin\theta }}\left[ \frac{h}{\ell }=sin\theta  \right]\] \[\operatorname{t}=0\sqrt{\frac{2\ell }{g\,sin\theta }\frac{h}{sin\theta }}\] \[\operatorname{t}=\frac{1}{sin\theta }\sqrt{\frac{2h}{\operatorname{g}}}\]