A) 2.1V
B) 1.8V
C) 0.2V
D) \[1.2\,V\]
Correct Answer: C
Solution :
(c) 0.2V Here, Charge, \[2a=1.38\times {{10}^{-10}}m,\,r=10\times {{10}^{-10}}m\] As potential, \[V=\frac{p}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}=\frac{q\left( 2a \right)}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}\] \[=\frac{9\times {{10}^{9}}\times 1.6\times {{10}^{-19}}\times 1.38\times {{10}^{-10}}}{{{\left( 10\times {{10}^{-10}} \right)}^{2}}}=0.2\,V\]You need to login to perform this action.
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