The ratio of electrostatic and gravitational | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(a) Gravitational force ${F_G} = \frac{{G{m_e}{m_p}}}{{{r^2}}}$
${F_G} = \frac{{6.7 	imes {{10}^{ - 11}} 	imes 9.1 	imes {{10}^{ - 31}} 	imes 1.6

Vedclass · Accepted Answer

$2.36 	imes 10^{39}$

Vedclass · Answer

(a) Gravitational force ${F_G} = \frac{{G{m_e}{m_p}}}{{{r^2}}}$
${F_G} = \frac{{6.7 	imes {{10}^{ - 11}} 	imes 9.1 	imes {{10}^{ - 31}} 	imes 1.6 	imes {{10}^{ - 27}}}}{{{{(5 	imes {{10}^{ - 11}})}^2}}}= 3.9 	imes 10^{-47} \,N$
Electrostatic force ${F_e} = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{e^2}}}{{{r^2}}}$
${F_e} = \frac{{9 	imes {{10}^9} 	imes 1.6 	imes {{10}^{ - 19}} 	imes 1.6 	imes {{10}^{ - 19}}}}{{{{(5 	imes {{10}^{ - 11}})}^2}}} = 9.22 	imes 10^{-8}\, N$
So, $\frac{{{F_e}}}{{{F_G}}} = \frac{{9.22 	imes {{10}^{ - 8}}}}{{3.9 	imes {{10}^{ - 47}}}} = 2.36 	imes {10^{39}}$

Similar Questions

A negatively charged particle $p$ is placed, initially at rest, in $a$ constant, uniform gravitational field and $a$ constant, uniform electric field as shown in the diagram. What qualitatively, is the shape of the trajectory of the electron?

What is the net force on a $Cl^{-}$ placed at the centre of the bcc structure of $CsCl$

A $10\,\mu C$ charge is divided into two parts and placed at $1\,cm$ distance so that the repulsive force between them is maximum. The charges of the two parts are :

A charge $q$ is placed at the centre of the line joining two equal charges $Q$. The system of the three charges will be in equilibrium, if $q$ is equal to

A point charge $q_1$ exerts force $F$ upon another point charge $q_2$. If a third charge $q_3$ be placed near the charge $q_2$, then the force that charge $q_1$ exerts on the charge $q_2$ will be