Question

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

Answer 1

Answer: (B)

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 $