Question

Exercise 15.1 (a) Find the magnitude of the electric force between two protons separated by 1 femtometer (10^-15), approximately the distance between two protons in the nucleus of a helium atom. (b) If the protons were not held together by the strong nuclear force, what would be their initial acceleration due to the electric force between them

Answer 1

Answer:

(a) 230.4 N

(b) $a = 1.38\times10^{29} m/s^{2}$

Explanation:

Charge on a proton, q = $1.6\times10^{-19}C$

Distance between the two protons, d = 1 femto meter = $1\times10^{-15}m$

(a) By use of coulomb's law, the force between the two protons is given by

$F =\frac{k\times q\times q}{d^{2}}$

Where, k is the Coulombic constant = 9x 10^9 Nm^2/C^2

So, the force between them is given by

$F =\frac{9\times10^{9}\times 1.6\times10^{-19}\times 1.6\times10^{-19}}{\left (1\times10^{-15}  \right )^{2}}$

F = 230.4 N

(b) Acceleration is the ratio of force to the mass of proton.

the mass of proton, m = 1.67 x 10^-27 kg

So, acceleration

$a=\frac{230.4}{1.67\times10^{-27}}$

$a = 1.38\times10^{29} m/s^{2}$