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 :
Potential difference between two equipotential surfaces A and B, \[{{V}_{A}}-{{V}_{B}}=kq\left( \frac{1}{{{r}_{A}}}-\frac{1}{{{r}_{B}}} \right)\] \[=kq\left( \frac{{{r}_{B}}-{{r}_{A}}}{{{r}_{A}}{{r}_{B}}} \right)\] \[=\frac{kq{{t}_{1}}}{{{r}_{A}}{{r}_{B}}}\] or \[{{t}_{1}}=\frac{({{V}_{A}}-{{V}_{B}}){{r}_{A}}{{r}_{B}}}{kq}\] or \[{{t}_{1}}\propto {{r}_{A}}{{r}_{B}}\] Similarly, \[{{t}_{2}}\propto {{r}_{B}}{{r}_{C}}\] Since \[{{r}_{A}}<{{r}_{B}}<{{r}_{C}},\]therefore \[{{r}_{A}}{{r}_{B}}<{{r}_{B}}{{r}_{C}}\] \[\therefore \] \[{{t}_{1}}<{{t}_{2}}\]
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