Question
Question: let us assume that the charge of a proton and an electron is varying slightly. One of the charges is...
let us assume that the charge of a proton and an electron is varying slightly. One of the charges is −e and the other charge is (e+Δe). When the resultant of the electrostatic force and gravitational force between two hydrogen atoms kept at a distance d (much greater than atomic size) will be zero, then Δe is of the order of [Given mass of hydrogen is mh=1.67×10−27kg]
& A{{.10}^{-20}}C \\\ & B{{.10}^{-23}}C \\\ & C{{.10}^{-37}}C \\\ & D{{.10}^{-47}}C \\\ \end{aligned}$$Solution
first of all find the net charge of the hydrogen atom. The gravitational force can be found by taking the ratio of the product of the gravitational constant and the square of mass of the hydrogen atom to the square of the radius of the atom. Electrostatic force is found by taking the product of the coulomb’s constant and the square of the charge to the square of the radius of the atom. Compare both these equations.
Complete step by step answer:
the mass of the hydrogen atom can be shown as,
mh=1.67×10−27kg
The net charge on the hydrogen atom will be,
q=(e+Δe)−e=Δe
Let the distance between these atoms be d.
Now we can equate the gravitational force to the electrostatic force. This time we get,
d2Gmh2=d2K(Δe)2
Substituting the values in it will give,
d2(6.67×10−11)×(1.67×10−27)2=d2(9×109)(Δe)2
The denominators can be cancelled. The value of the net charge will be given as,
Δe=2.06×10−37C
So, the correct answer is “Option C”.
Note: The electrostatic force is otherwise called the Coulomb interaction or Coulomb force. It may be the attractive or repulsive force between two electrically charged bodies. As we all know the similar charges will repel each other and also the opposite charges can attract each other. Coulomb's law is helpful in measuring the strength of the force between two charges.