Question

Two electrons are placed at a distance of 1fermi (fertometer). Calculate the gravitational force of attraction between the. Given: mass of electron 9.1*10^31kg and 1 fermi = 10^15 m).​

Answer 1

Answer:

Gravitational force, $F=5.52\times 10^{-41}\ N$

Explanation:

It is given that,

Mass of electrons, $m_1=m_2=9.1\times 10^{-31}\ kg$

Distance between electrons, $d=1\ fermi=10^{-15}\ m$

We need to find the gravitational force of attraction between the two electrons. The formula for calculating the gravitational force is given by :

$F=G\dfrac{m_1m_2}{r^2}$

$F=6.67\times 10^{-11}\times \dfrac{(9.1\times 10^{-31})^2}{(10^{-15})^2}$

$F=5.52\times 10^{-41}\ N$

So, the force of attraction between two electrons is $5.52\times 10^{-41}\ N$. Hence, this is the required solution.

Answer 2

Explanation:

Answer:

Gravitational force, F=5.52\times 10^{-41}\ NF=5.52×10

−41

N

Explanation:

It is given that,

Mass of electrons, m_1=m_2=9.1\times 10^{-31}\ kgm

1

=m

2

=9.1×10

−31

kg

Distance between electrons, d=1\ fermi=10^{-15}\ md=1 fermi=10

−15

m

We need to find the gravitational force of attraction between the two electrons. The formula for calculating the gravitational force is given by :

F=G\dfrac{m_1m_2}{r^2}F=G

r

2

m

1

m

2

F=6.67\times 10^{-11}\times \dfrac{(9.1\times 10^{-31})^2}{(10^{-15})^2}F=6.67×10

−11

×

(10

−15

)

2

(9.1×10

−31

)

2

F=5.52\times 10^{-41}\ NF=5.52×10

−41

N

So, the force of attraction between two electrons is 5.52\times 10^{-41}\ N5.52×10

−41

N . Hence, this is the required solution.