A) \[5.76\times {{10}^{11}}N/C\]
B) \[1.44\times {{10}^{11}}N/C\]
C) \[2.828\times {{10}^{11}}N/C\]
D) zero
Correct Answer: B
Solution :
The relation for intensity of electric field is: \[E=\frac{1}{4\pi \varepsilon 0}\times \frac{q}{{{r}^{2}}}\] Here: \[\frac{1}{4\pi {{\varepsilon }_{0}}}=9\times {{10}^{9}}\] \[r=1\overset{\text{o}}{\mathop{\text{A}}}\,=1\times {{10}^{-10}}m\] \[q=1.6\times {{10}^{-19}}C\] \[E=\frac{9\times {{10}^{9}}\times 1.6\times {{10}^{-19}}}{{{(1\times {{10}^{-10}})}^{2}}}\] \[=1.44\times {{10}^{11}}N/C\]
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