Question

An electron is separated from a proton through a distance of 0.53 Aº
Calculate electric field at the location of electron​

Answer 1

It may be like this:

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Answer 2

Answer:

Electric field at the location of electron​ is $E=5.1 \times 10^{11} \mathrm{~N} / \mathrm{m}$

Explanation:

Coulomb's Law is applied to a single charge. When a point charge is separated from the charge, it will be subjected to a force.

The electric field at this location is described by the following equation: This is the unit vector along the direction of the electric field at a distance from a point charge.

Given information

$K=9 \times 10^{9}$

$Q=1.6 \times 10^{-19}$

$r=0.53 \times 10^{-10}$

Force of the particle

$\begin{aligned}&F=q E \\&\Longrightarrow F=3.2 \times 10^{-19} \times 15 \times 10^{4} \\&\Longrightarrow F=4.8 \times 10^{-14} \mathrm{~N}\end{aligned}$

If the electric field due to a point charge is positive, it is pointed away from the charge; if it is negative, it is pointed towards the charge, and its magnitude decreases as the square of the inverse distance from the charge increases.

Now for Electric field .

Applying

$E=\frac{K Q}{r^{2}}$

$E=\frac{9 \times 10^{9} \times 1.6 \times 10^{-19}}{\left(0.53 \times 10^{-10}\right)^{2}}$

$E=\frac{9 \times 1.6}{0.53 \times 0.53} \times 10^{9-19+20}$

$E=5.1 \times 10^{11} \mathrm{~N} / \mathrm{m}$