Question

The mass of an electron is $ 9.11\times {{10}^{-31}}kg $ , that of proton is $ 1.67\times {{10}^{-27}}kg $ . Find the ratio of the electric force and the gravitational force exerted by the proton on the electron. $ \left( \dfrac{{{F}_{e}}}{{{F}_{g}}} \right) $ .

Answer 1

Hint :Calculate the gravitational force and the electric force between the proton and the electron by assuming some fixed distance between them. The magnitude of charge on each of the particles is $ e=1.6\times {{10}^{-19}}C $ . Then divide the two to find the ratio.
 $ {{F}_{g}}=\dfrac{G{{m}_{1}}{{m}_{2}}}{{{d}^{2}}} $ , where $ {{F}_{g}} $ is gravitational force between masses $ {{m}_{1}} $ and $ {{m}_{2}} $ separated by a distance d. G is the gravitational constant.
 $ {{F}_{e}}=\dfrac{K{{q}_{1}}{{q}_{2}}}{{{d}^{2}}} $ , where $ {{F}_{e}} $ is electrical force between charges $ {{q}_{1}} $ and $ {{q}_{2}} $ separated by a distance d. K is the Coulomb’s constant.

Complete Step By Step Answer:
Let us assume that is a proton and electron fixed and separated by a distance d in vacuum.
Let us first find the gravitational force exerted by the proton on the electron. It is given that the mass of the proton is equal to $ {{m}_{p}}=1.67\times {{10}^{-27}}kg $ and the mass of an electron is equal to $ {{m}_{e}}=9.11\times {{10}^{-31}}kg $ .
Now, the gravitational force between the two particles is equal to $ {{F}_{g}}=\dfrac{G{{m}_{p}}{{m}_{e}}}{{{d}^{2}}} $ …. (i)
The value of G in SI units is equal to $ G=6.67\times {{10}^{-11}}N{{m}^{2}}k{{g}^{-2}} $
Substitute the values of $ {{m}_{p}} $ , $ {{m}_{e}} $ and G in equation (i).
 $ \Rightarrow {{F}_{g}}=\dfrac{\left( 6.67\times {{10}^{-11}} \right)\left( 1.67\times {{10}^{-27}} \right)\left( 9.11\times {{10}^{-31}} \right)}{{{d}^{2}}} $ …. (ii).
Let us now calculate the electric force exerted by the proton on the electron.
The magnitude of charge on an electron and a proton is equal to $ e=1.6\times {{10}^{-19}}C $ .
Therefore, the electric force between the two charged particles is equal to $ {{F}_{e}}=\dfrac{Ke.e}{{{d}^{2}}} $ …. (iii)
The value of K, in SI units is equal to $ K=9\times {{10}^{9}}N{{m}^{2}}{{C}^{-2}} $ .
Substitute the value of K and e in equation (iii).
With this we get that
 $ {{F}_{e}}=\dfrac{\left( 9\times {{10}^{9}} \right)\left( 1.6\times {{10}^{-19}} \right)\left( 1.6\times {{10}^{-19}} \right)}{{{d}^{2}}} $ …. (iv)
Now, divide (iv) by (ii).
Then,
 $ \Rightarrow \dfrac{{{F}_{e}}}{{{F}_{g}}}=\dfrac{\dfrac{\left( 9\times {{10}^{9}} \right)\left( 1.6\times {{10}^{-19}} \right)\left( 1.6\times {{10}^{-19}} \right)}{{{d}^{2}}}}{\dfrac{\left( 6.67\times {{10}^{-11}} \right)\left( 1.67\times {{10}^{-27}} \right)\left( 9.11\times {{10}^{-31}} \right)}{{{d}^{2}}}} $
 $ \Rightarrow \dfrac{{{F}_{e}}}{{{F}_{g}}}=2.27\times {{10}^{39}} $ .

Note :
With this question, we can understand the difference in the magnitude of the electric force and the gravitational force between an electron and a proton. We get that the electric force between the two particles is much much greater than the gravitational force between the two. Therefore, we can neglect the gravitational force of attraction when we talk about forces between an electron and a proton.
Note that the electric force is an attractive force in this case.