Suggested languages for you:

Americas

Europe

Q22PE

Expert-verified
Found in: Page 664

### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# At what distance is the electrostatic force between two protons equal to the weight of one proton?

The electrostatic force between two protons equal to the weight of one proton, when the two protons are placed at a separation of 0.119 m.

See the step by step solution

## Step 1: Given data

The force of repulsion between two protons is equal to the weight experienced by a proton.

## Step 2: Weight

Weight of an object is defined as the force by which the Earth attracts the body. The expression for the weight is given as,

$F=mg$

Here, F is the weight of the object, m is the mass of the object and g is the acceleration due to gravity.

## Step 3: Calculating distance

The electrostatic force between two protons is,

${F}_{e}=\frac{K{q}^{2}}{{r}^{2}}----------\left(1.1\right)$

Here, is the electrostatic force constant $\left(K=9×{10}^{9}N-{m}^{2}/{C}^{2}\right)$ , q is the charge on protons $\left(q=1.6×{10}^{-19}C\right)$ , and r is the separation between the protons.

The weight of one proton is,

$F={m}_{p}g----------\left(1.2\right)$

Here, mp is the mass on a proton $\left({m}_{p}=1.67×{10}^{-27}kg\right)$ , and g is the acceleration due to gravity $\left(g=9.8m/{s}^{2}\right)$ .

Since, the electrostatic force between the protons equals the weight of one proton.

Equating equation (1.1) and (1.2),

${m}_{p}g=\frac{K{q}^{2}}{{r}^{2}}$

The expression for the separation between the protons is given as,

$r=\sqrt{\frac{K{q}^{2}}{{m}_{p}g}}$

Substituting all known values,

$r=\sqrt{\frac{\left(9×{10}^{9}N-{m}^{2}/{C}^{2}\right)×{\left(1.6×{10}^{-19}C\right)}^{2}}{\left(1.67×{10}^{-27}kg\right)×\left(9.8m/{s}^{2}\right)}}\phantom{\rule{0ex}{0ex}}=0.119m$

Hence, at a distance of 0.119 m the electrostatic force between two protons equal to the weight of one proton.