A) \[1.8\times {{10}^{5}}V\text{ and }-5.625\times {{10}^{6}}V/m\]
B) 0 V and 0 V/m
C) \[2.25\times {{10}^{5}}V\text{ and }-5.625\times {{10}^{6}}V/m\]
D) \[2.25\times {{10}^{5}}V\text{ and 0 V/m}\]
Correct Answer: C
Solution :
\[q=1\mu C=1\times {{10}^{-6}},r=4cm=4\times {{10}^{-2}}m\] Potential \[V=\frac{kg}{r}=\frac{9\times {{10}^{8}}\times {{10}^{-6}}}{4\times {{10}^{-2}}}\] \[=2.25\times {{10}^{5}}V.\] 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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