A) \[\text{5}.\text{76}\times \text{l}{{0}^{\text{11}}}\text{ N}/\text{C}\]
B) \[\text{1}.\text{44}\times \text{l}{{0}^{\text{11}}}\text{ N}/\text{C}\]
C) \[\text{2}.\text{828}\times \text{l}{{0}^{\text{11}}}\text{ N}/\text{C}\]
D) zero
Correct Answer: B
Solution :
The intensity of electric field \[(\vec{E})\] due to charge \[E=\frac{1}{4\pi \,{{\varepsilon }_{0}}\,{{r}^{2}}}\frac{q}{{{r}^{2}}}\] Given, \[r=1\,\overset{o}{\mathop{A}}\,=1\times {{10}^{-10}}\,m\], \[q=1.6\times {{10}^{-19}}C\] \[\therefore \] \[E=9\times {{10}^{9}}\times \frac{1.6\times {{10}^{-19}}}{{{(1\times {{10}^{-10}})}^{2}}}\] \[=1.44\times {{10}^{11}}N/C\]You need to login to perform this action.
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