question_answer

Deuteron and alpha particle in air are at separation \[1\overset{\text{o}}{\mathop{\text{A}}}\,\] The magnitude of electric field intensity on a-particle due to deuteron is

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,   \[r=1\,\overset{o}{\mathop{A}}\,=1\times {{10}^{-10}}\,m\],                 \[q=1.6\times {{10}^{-19}}C\] \[\therefore \]  \[E=9\times {{10}^{9}}\times \frac{1.6\times {{10}^{-19}}}{{{(1\times {{10}^{-10}})}^{2}}}\]                 \[=1.44\times {{10}^{11}}N/C\]