A) \[1.7\times {{10}^{-11}}\,N/C\]
B) \[5.0\times {{10}^{-11}}\,N/C\]
C) \[5.5\times {{10}^{-11}}\,N/C\]
D) 56 N/C
Correct Answer: C
Solution :
\[E=\frac{F}{q}=\frac{mg}{e}\] \[=\frac{9\times {{10}^{-31}}\times 9.8}{1.6\times {{10}^{-19}}}=5.5\times {{10}^{-11}}N/C\]You need to login to perform this action.
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