The nucleus of the helium atom contains two protons that are separated by about 3.8 multiplied by 10^-15 m. How do you find the magnitude of the electrostatic force that each proton exerts on the other?

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The nucleus of the helium atom contains two protons that are separated by about 3.8 multiplied by 10^-15 m. How do you find the magnitude of the electrostatic force that each proton exerts on the other?

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Answer 1 · 2015-12-30T08:39:13

$F = 15.956 N$ $F=15.956N$

Explanation:

Data:
Charge on each proton $= q_{1} = q_{2} = q = 1.6 \cdot 10^{- 19} C$ $=q_1=q_2=q=1.6*10^-19C$
Distance between charges $= r = 3.8 \cdot 10^{- 15} m$ $=r=3.8*10^-15m$
Sol:-
Electrostatic force between two charges is given by
$F = \frac{k q_{1} q_{2}}{r^{2}}$ $F=(kq_1q_2)/r^2$
Where $k$ $k$ is the constant and its value is $9 \cdot 10^{9}$ $9*10^9$
Here $q_{1} = q_{2} = q$ $q_1=q_2=q$
$\Rightarrow F = \frac{k q q}{r^{2}} \Rightarrow F = \frac{k q^{2}}{r^{2}}$ $implies F=(kqq)/r^2=>F=(kq^2)/r^2$
$\Rightarrow F = \frac{9 \cdot 10^{9} {(1.6 \cdot 10^{- 19})}^{2}}{{(3.8 \cdot 10^{- 15})}^{2}} = 15.956 N$ $implies F=(9*10^9(1.6*10^-19)^2)/(3.8*10^-15)^2=15.956N$
$\Rightarrow F = 15.956 N$ $implies F=15.956N$

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The nucleus of the helium atom contains two protons that are separated by about 3.8 multiplied by 10^-15 m. How do you find the magnitude of the electrostatic force that each proton exerts on the other?

1 Answer

Explanation:

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