Question

Deuteron and alpha particle in air are at separation $1\stackrel{0}{A}$ . 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: $E=\frac{1}{4\pi \epsilon 0}\times \frac{q}{r^2}$ Here: $\frac{1}{4\pi {\epsilon }_0}=9\times {10}^9$ $r=1\stackrel{\text{o}}{A}=1\times {10}^{-10}m$ $q=1.6\times {10}^{-19}C$ $E=\frac{9\times {10}^9\times 1.6\times {10}^{-19}}{{(1\times {10}^{-10})}^2}$ $=1.44\times {10}^{11}N/C$