question_answer

Two capacitors of capacitances \[\text{3}\,\text{ }\!\!\mu\!\!\text{ F}\]and \[6\,\text{ }\!\!\mu\!\!\text{ F}\]are charged to a potential of 12 V each. They are now connected to each other, with the positive plate to each joined to the negative plate to the other. The potential difference across each will be:

A)  4V                                         

B)  6V

C)  zero                                     

D)  3 V

Correct Answer: A

Solution :

\[{{F}_{1}}=(F)\]one force \[{{F}_{2}}=(2F)\]second force Let \[\theta \]is angle between \[{{F}_{1}}\]and \[{{F}_{2}}.\] Here angle made by resultant \[{{F}_{R}}\]is \[\alpha \]with \[{{F}_{1}}\] \[\therefore \]  \[\alpha =90\] \[\tan \alpha =\frac{{{F}_{2}}\sin \theta }{{{F}_{1}}+{{F}_{2}}\cos \theta }=\alpha \] \[\frac{{{F}_{2}}\sin \theta }{{{F}_{1}}+{{f}_{2.}}\cos \theta }=\frac{1}{0}\] \[{{F}_{1}}+{{F}_{2}}\cos \theta =0\] \[2F\cos \theta =-F\] \[\cos \theta =-1/2\] \[\theta =120\]