Question

Free charges do not remain stationary when close together. To illustrate this, calculate the magnitude of the instantaneous acceleration in, meters per second squared, of two isolated protons separated by 2.5 nm.

Answer 1

Answer:

a=2.304×10¹⁶m/s²

Explanation:

Given data

Distance d=2.5 nm=2,5×10⁻⁹m

Mass of proton m=1.6×10⁻²⁷kg

charge of proton q=1.6×10⁻¹⁹C

To find

acceleration a

Solution

Apply the Coulombs Law

$F=k\frac{q_{1}q_{2} }{r^{2} }$

Where k is coulombs constant (k=9×10⁹Nm²/C²)

q=q₁=q₂

r=d

So

$F=k\frac{|q^{2} |}{d^{2} }\\ as \\F=ma\\ma=k\frac{|q^{2} |}{d^{2} }\\a=\frac{k}{m} \frac{|q^{2} |}{d^{2} }\\a=\frac{(9*10^{9} )*(1.6*10^{-19} )^{2} }{(1.6*10^{-27} )*(2.5*10^{-9} )^{2} }\\ a=2.304*10^{16}m/s^{2}$