A) \[1.8\times {{10}^{5}}\] and \[-5.625\times {{10}^{6}}\,V/m\]
B) \[0\operatorname{V}\] and \[0\operatorname{V}/\operatorname{m}\]
C) \[2.25\times {{10}^{5}}\operatorname{V}\] and \[-5.625\times {{10}^{6}}\operatorname{V}/\operatorname{m}\]
D) \[2.25\times {{10}^{5}}\operatorname{V}\] and V/m
Correct Answer: C
Solution :
\[q=1\mu C=1\times {{10}^{-6}}C\] \[r=4\,cm=4\times {{10}^{-2}}m\] Potential \[V=\frac{kq}{r}\] \[=\frac{9\times {{10}^{9}}\times {{10}^{-6}}}{4\times {{10}^{-2}}}\] \[=2.25\times {{10}^{5}}V.\] Induced electric field \[E=\frac{kq}{{{r}^{2}}}\] \[=\frac{9\times {{10}^{9}}\times 1\times {{10}^{-6}}}{16\times {{10}^{-4}}}=-5.625\times {{10}^{6}}\,V/m\]
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