Question

The ${F_g}$ and ${F_e}$ represent gravitational and electrostatic force respectively between electrons situated at a distance $0.1\,m$ . ${F_g}/{F_e}$ is of the order ?
A. ${\text{1}}{{\text{0}}^{ - 41}}$
B. ${\text{1}}{{\text{0}}^{ - 45}}$
C. ${\text{1}}{{\text{0}}^{40}}$
D. ${\text{1}}{{\text{0}}^{ - 42}}$

Answer 1

To solve this sort of question we use the concepts of gravity and electrostatic force.Then we substitute the given values within the formulae of gravity and electrostatic force.Finally we take ratio of those to seek out the order.

Formula used:
Gravitational force is given by,
${F_g} = \dfrac{{G{m_1}{m_2}}}{{{r^2}}}$
Where $G$ refers to the universal gravitational constant , adequate to $6.67 \times {10^{ - 11}}\dfrac{{{m^3}}}{{kg.{s^2}}}$
Electrostatic force is given by,
${F_e} = \dfrac{{k{q_1}{q_2}}}{{{r^2}}}$
Where, ${q_1}$ and ${q_2}$ are the two charges placed with a distance ‘$r$’ between them.

Complete step by step answer:
The gravity is that which attracts any two objects with mass. We call this force attractive because it always tries to tug the masses together, but it never pushes them apart. In fact, every object, including we (a human body), is pulling on every other object during this complete universe! This is often simply nothing but Newton's Universal Law of Gravitation.Admittedly, we don't have a huge mass, and so, we're not pulling much on those other objects. Moreover, objects that are really far aside from one another don't either pull on one another noticeably. Nevertheless, there's the force, which we will calculate.

The equation for Newton gravity is given by,
${F_g} = \dfrac{{G{m_1}{m_2}}}{{{r^2}}}$
Electrostatic force is additionally called coulomb force, deals with static electric charges or charges at rest. Coulomb’s law states that the force of attraction or repulsion between two-point charges is directly proportional to the merchandise of the fees and inversely proportional to the square of the space between them. The directions of forces are along the road joining the two charges.

Consider ${q_1}$ and ${q_2}$as two charges placed with a distance ‘r’ between them.Then, consistent with Coulomb’s law,
${F_e} = \dfrac{{k{q_1}{q_2}}}{{{r^2}}}$
$\Rightarrow {F_e} = \dfrac{1}{{4\pi { \in _0}}}\dfrac{{{q_1}{q_2}}}{{{r^2}}}$.....($\dfrac{1}{{4\pi { \in _0}}} = 9 \times {10^9}N{m^2}{c^{ - 2}}$)
$\Rightarrow {F_e} = 9 \times {10^9}\dfrac{{{q_1}{q_2}}}{{{r^2}}}$

Now we are given to seek out the order of ratio $\dfrac{{{F_g}}}{{{F_e}}}$ between two electrons separated at a distance $0.1\,m$.
${m_1} = {m_2} = m$.......(mass of electron $m = 9.1 \times {10^{ - 31}}kg$)
$\Rightarrow {q_1} = {q_2} = q$.....(charge of electron $q = 1.6 \times {10^{ - 19}}C$)
$\Rightarrow d = 0.1m$

\Rightarrow \dfrac{{{F_g}}}{{{F_e}}} = \dfrac{1}{{9 \times {{10}^9}}} \times 6.67 \times {10^{ - 11}} \times \dfrac{{{{\left( {9.1 \times {{10}^{ - 31}}} \right)}^2}}}{{{{\left( {1.6 \times {{10}^{ - 19}}} \right)}^2}}}$$ $$\therefore \dfrac{{{F_g}}}{{{F_e}}} = 2.397 \times {10^{ - 45}}$$ Therefore the ratio is of order of $${10^{ - 45}}$$. **Thus,option B is the correct answer.** **Note:** There's a difference between the electrostatic force and gravity The electrostatic force is for charges and gravity is for masses. The forces are often positive or negative but mass can't be negative.