Suggested languages for you:

Americas

Europe

Q22PE

Expert-verifiedFound in: Page 664

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.

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

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.

The electrostatic force between two protons is,

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

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

The weight of one proton is,

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

Here, *m _{p}* is the mass on a proton $\left({m}_{p}=1.67\times {10}^{-27}kg\right)$ , and

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,

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

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

94% of StudySmarter users get better grades.

Sign up for free