A) \[{{t}_{1}}\,\,=\,\,{{t}_{2}}\]
B) \[{{t}_{1}}\,\,>\,\,{{t}_{2}}\]
C) \[{{t}_{1}}\,\,<\,\,{{t}_{2}}\]
D) \[{{t}_{1}}\,\,\le \,\,{{t}_{2}}\]
Correct Answer: C
Solution :
\[{{V}_{A}} -{{V}_{B}} = potential difference between\] \[=\,\,\,kq\left( \frac{1}{{{r}_{A}}}-\frac{1}{{{r}_{B}}} \right)\,\,=\,\,kq\left( \frac{{{r}_{B}}-{{r}_{A}}}{{{r}_{A}}{{r}_{B}}} \right)\] \[=\,\,\,\frac{kq{{t}_{1}}}{{{r}_{A}}{{r}_{B}}}\] \[\Rightarrow \,\,\,\,\,\,\,{{t}_{1}}=\frac{({{V}_{A}}-{{V}_{B}}){{r}_{A}}{{r}_{B}}}{kq}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,{{t}_{1}}\propto {{r}_{A}}{{r}_{B}}\] Similarly \[{{t}_{2}}\propto {{r}_{B}}{{r}_{C}}\] Since \[{{r}_{A}}<{{r}_{B}}<{{r}_{C}}\,\,\,\Rightarrow \,\,\,{{r}_{A}}{{r}_{B}}<{{r}_{B}}{{r}_{C}}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{t}_{1}}<{{t}_{2}}\]You need to login to perform this action.
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