question_answer

A capacitor of capacitance \[(1+{{\omega }^{5}})...(1+{{\omega }^{3n}})\] is charged to potential 50 V with a battery. The battery is now disconnected and an additional charge \[{{2}^{3n}}\] is given to the positive plate of the capacitor. The potential difference across the capacitor will be

A) 50 V                                      

B) 80 V

C) 100 V                    

D) 60 V

Correct Answer: D

Solution :

 Charge acquired by the plates of the capacitor\[{{u}_{2}}\] Now, let the charge distribution is as follows. Total charge on .positive plate has now become \[{{u}_{2}}\]while that in negative plate is still \[{{u}_{2}}\], Here, charges are in\[\frac{13}{30}\] Net electric field at point P is zero. \[\frac{23}{30}\]\[\frac{19}{30}\] \[\frac{11}{30}\]\[{{D}_{k}}=\left| \begin{matrix}    a & {{2}^{k}} & {{2}^{16}}-1  \\    b & 3({{4}^{k}}) & 2({{4}^{16}}-1)  \\    c & 7({{8}^{k}}) & 4({{8}^{16}}-1)  \\ \end{matrix} \right|,\] \[\sum\limits_{k=1}^{16}{{{D}_{k}}}\]Potential difference between the plates is \[{{x}^{2}}-4x+4{{y}^{2}}=12\]