A) \[2.88\times {{10}^{11}}\text{N/C}\]
B) \[1.44\times {{10}^{11}}\text{N/C}\]
C) \[5.76\times {{10}^{11}}\text{N/C}\]
D) zero
Correct Answer: B
Solution :
Charge on \[_{1}{{H}^{2}}(Deuteron)=e\] So, intensity at any point due to deuteron particle is \[E=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{e}{{{r}^{2}}}\] \[=\frac{9\times {{10}^{9}}\times 1.6\times {{10}^{-19}}}{{{(1\times {{10}^{-10}})}^{2}}}\] \[=1.44\times {{10}^{11}}N/C\]You need to login to perform this action.
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