Question

Deuteron and alpha particle in air are at separation 1ampere . the magnitude of electric field intensity on alpha particle due to deutron is​

Answer 1

hey mate I am answering your question please comment me that how was my answer

answer :
$1.44 \times {10}^{11} {Nc}^{ - 1}$
solution :
charge on deuteron= +e =
$16 \times {10}^{ - 19}$
$E = \frac{1}{4\pi \: e_{0} } \times \frac{q}{ {r}^{2} }$
$\frac{1}{4\pi \: e_{0} } \times \frac{q}{ {r}^{2} } = 9 \times {10}^{9} \times \frac{16x {10}^{ - 19} } {({1x {10}^{ - 10} )}^{2} }$
$9 \times {10}^{9} \times \frac{16x {10}^{ - 19} } {({1x {10}^{ - 10} )}^{2} } = 1.44 \times {10}^{11} {Nc}^{ - 1}$
hope it will helpful for you