Question

Bare free charges do not remain stationary when close together. To illustrate this, calculate the acceleration of two isolated protons separated by 2.00 nm (a typical distance between gas atoms). Explicitly show how you follow the steps in the Problem-Solving Strategy for electrostatics.

Answer 1

Answer:

Acceleration, $a=3.6\times 10^{16}\ m/s^2$

Explanation:

It is given that, two isolated protons are separated by 2 nm. The force due to charged particles is given by :

$F_e=\dfrac{kq^2}{d^2}$

Force due to mass of proton, $F_g=ma$

$ma=\dfrac{kq^2}{d^2}$

$a=\dfrac{kq^2}{md^2}$

$a=\dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{1.6\times 10^{-27}\times (2\times 10^{-9})^2}$

$a=3.6\times 10^{16}\ m/s^2$

So, the acceleration of two isolated protons is $3.6\times 10^{16}\ m/s^2$. Hence, this is the required solution.