Question

Find the electrostatic force between two protons placed in free space separated by a distance 20 cm

Answer 1

$q = 1.6 \times {10}^{ - 19}c\\ r = .20m \\ k = 9 \times {10}^{9}\frac{n {m}^{2} }{ {c}^{2} }f = \frac{k {q}^{2} }{ {r}^{2} } \\ f = 5.76 \times {10}^{ - 27}$

Answer 2

The electrostatic force between two protons placed in free space separated by a distance 20 cm is $5.76\times 10^{-27}\ N$.  

Explanation:

Given that,

Distance between two protons, r = 20 cm = 0.2 m

Charges on proton, $q_1=q_2=1.6\times 10^{-19}\ C$

If F is the electric force between two charges. It is given by the electrostatic force of attraction or repulsion. It is given by :

$F=\dfrac{kq_1q_2}{d^2}$

$F=\dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{(0.2)^2}$

$F=5.76\times 10^{-27}\ N$

So, the electrostatic force between two protons placed in free space separated by a distance 20 cm is $5.76\times 10^{-27}\ N$.  

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