Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11

# Find the Ratio of the Magnitude of the Electric Force to the Gravitational Force Acting Between Two Protons. - Physics

Sum

Find the ratio of the magnitude of the electric force to the gravitational force acting between two protons.

#### Solution

Charge of the proton, q = $1 . 6 \times {10}^{- 19} C$

Mass of the proton = $1 . 67 \times {10}^{- 27} \text{ kg }$

Let the distance between two protons be r.
Coulomb force (electric force) between the protons is given by

$f_e = \frac{1}{4\pi \in_0} \times \frac{q^2}{r^2}$

$= \frac{9 \times {10}^9 \times (1 . 6 )^2 \times {10}^{- 38}}{r^2}$

Gravitational force between the protons is given by

$f_g = \frac{\text{ G m}^2}{r^2}$

$= \frac{6 . 67 \times {10}^{- 11} \times (1 . 67 \times {10}^{- 27} )^2}{r^2}$

On dividing , $f_e \text{ by } f_g$  , We get :

$\frac{f_e}{f_g} = \frac{1}{4\pi \in_0} \times \frac{q^2}{r^2} \times \frac{r^2}{\text{ G m} ^2}$

$= \frac{9 \times {10}^9 \times 1 . 6 \times 1 . 6 \times {10}^{- 38}}{6 . 67 \times {10}^{- 11} \times 1 . 67 \times 1 . 67 \times {10}^{- 54}}$

$= \frac{9 \times (1 . 6 )^2 \times {10}^{- 29}}{6 . 67 \times (1 . 67 )^2 \times {10}^{- 65}}$

$= 1 . 24 \times {10}^{36}$

Concept: Work Done by a Constant Force and a Variable Force
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#### APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 1
Chapter 4 The Forces
Exercise | Q 10 | Page 63
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