VMMC VMMC Medical Solved Paper-2005

  • question_answer
    Deuteron and alpha particle in air are at separation\[1\overset{0}{\mathop{A}}\,\]. The magnitude of electric field intensity on \[\alpha \]-particle due to deuteron is:

    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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