A) 780 K
B) \[1.32\times {{10}^{5}}\,K\]
C) \[7.84\times {{10}^{4}}K\]
D) \[1900\,K\]
Correct Answer: C
Solution :
\[q'=(Q-q)\] where, \[\therefore \] \[F=\frac{1}{r\pi {{\varepsilon }_{0}}}\frac{Q(Q-q)}{{{R}^{2}}}\] \[\frac{dF}{dq}=0\] \[\frac{d}{dq}\,\left[ \frac{Kq(Q-q)}{{{R}^{2}}} \right]=0\] \[\Rightarrow \] \[[Q-2q]=0\] \[q=\frac{Q}{2}\] of H atom \[U=\frac{{{Q}^{2}}}{2C}\] \[C'=KC\] \[U'=\frac{{{Q}^{2}}}{2C'}=\frac{1}{2}\,\frac{{{Q}^{2}}}{KC}\]
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