Question

Calculate the intensity of electric feild at a point 4 mm away from an alpha particle.

Answer 1

Explanation:

In this

way we can solve the problem for you my friend

Answer 2

Given ,

Distance = 4 mm

Charge (q) = 2e

We know that , the electric field at point P due to point charge is given by

$\sf \fbox{E = k \frac{q}{ {(r)}^{2} } }$

Thus ,

$\sf \mapsto E = \frac{9 \times {(10)}^{9} \times 2 \times 1.6 \times {(10)}^{ - 19} }{ {(4 \times {(10)}^{ - 3})}^{2} } \\ \\ \sf \mapsto E =\frac{18 \times {(10)}^{ - 10} }{ {(10)}^{ - 5} } \\ \\\sf \mapsto E = 18 \times {(10)}^{ - 5} \: \: N/c$

$\therefore \sf \underline{The \: electric \: field \: is \: 18 \times {(10)}^{-5} \: \: N/c}$