The ratio of gravitational force and electros | vedclass

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

Question

(C) Given:Gravitational constant $G = 6.7 	imes 10^{-11} \, Nm^2/kg^2$Mass of an electron $m_e = 9.1 	imes 10^{-31} \, kg$Charge of an electron $e =

Vedclass · Accepted Answer

$24 	imes 10^{-44}$

Vedclass · Answer

(C) Given:Gravitational constant $G = 6.7 	imes 10^{-11} \, Nm^2/kg^2$Mass of an electron $m_e = 9.1 	imes 10^{-31} \, kg$Charge of an electron $e = 1.6 	imes 10^{-19} \, C$Coulomb constant $k = 9 	imes 10^9 \, Nm^2/C^2$Gravitational force between two electrons separated by distance $r$ is:$F_G = \frac{G m_e^2}{r^2} = \frac{6.7 	imes 10^{-11} 	imes (9.1 	imes 10^{-31})^2}{r^2}$Electrostatic repulsive force between two electrons is:$F_E = \frac{k e^2}{r^2} = \frac{9 	imes 10^9 	imes (1.6 	imes 10^{-19})^2}{r^2}$The ratio of gravitational force to electrostatic force is:$\frac{F_G}{F_E} = \frac{G m_e^2}{k e^2} = \frac{6.7 	imes 10^{-11} 	imes (9.1 	imes 10^{-31})^2}{9 	imes 10^9 	imes (1.6 	imes 10^{-19})^2}$Calculating the values:$\frac{F_G}{F_E} \approx \frac{6.7 	imes 82.81 	imes 10^{-11} 	imes 10^{-62}}{9 	imes 2.56 	imes 10^9 	imes 10^{-38}}$$\frac{F_G}{F_E} \approx \frac{554.827 	imes 10^{-73}}{23.04 	imes 10^{-29}} \approx 24.08 	imes 10^{-44}$Thus,the ratio is approximately $24 	imes 10^{-44}$.

The ratio of gravitational force and electrostatic repulsive force between two electrons is approximately (gravitational constant $= 6.7 \times 10^{-11} \, Nm^2/kg^2$,mass of an electron $= 9.1 \times 10^{-31} \, kg$,charge on an electron $= 1.6 \times 10^{-19} \, C$).

Similar Questions

$ABC$ is a right-angled triangle in which $AB = 3\,cm$ and $BC = 4\,cm$. And $\angle ABC = \pi / 2$. The three charges $+15\,e.s.u.$,$+12\,e.s.u.$,and $-20\,e.s.u.$ are placed respectively on $A$,$B$,and $C$. The force acting on $B$ is.......$dynes$.

Two charges of magnitude $5\, nC$ and $-2\, nC$ are placed at points $(2\, cm, 0, 0)$ and $(x\, cm, 0, 0)$ respectively in a region of space where there is no other external field. If the electrostatic potential energy of the system is $-0.5\,\mu J$,find the value of $x$ in $cm$.

Two point charges $+8 q$ and $-2 q$ are located at $x=0$ and $x=L$ respectively. The location of a point on the $x$-axis from the origin,at which the net electric field due to these two point charges is zero is

Two charges each equal to $2\,\mu C$ are $0.5\,m$ apart. If both of them exist inside vacuum,then the force between them is.......$N$

Vedclass Test Series

Exam Paper Generator

Online Exam Module