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 \[{{x}^{2}}-4x+4{{y}^{2}}=12\]You need to login to perform this action.
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