The acceleration of an electron due to the mu | vedclass

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

Question

(D) The electrostatic force between an electron and a proton is given by Coulomb's Law: $F = \frac{1}{4 \pi \varepsilon_{0}} \frac{e^{2}}{r^{2}}$.Acco

Vedclass · Accepted Answer

$10^{22} \; m/s^{2}$

Vedclass · Answer

(D) The electrostatic force between an electron and a proton is given by Coulomb's Law: $F = \frac{1}{4 \pi \varepsilon_{0}} \frac{e^{2}}{r^{2}}$.According to Newton's second law,the acceleration of the electron is $a_{e} = \frac{F}{m_{e}}$.Substituting the expression for force: $a_{e} = \frac{1}{4 \pi \varepsilon_{0}} \frac{e^{2}}{m_{e} r^{2}}$.Given $r = 1.6 \; \mathring{A} = 1.6 	imes 10^{-10} \; m$,$e = 1.6 	imes 10^{-19} \; C$,$m_{e} = 9 	imes 10^{-31} \; kg$,and $\frac{1}{4 \pi \varepsilon_{0}} = 9 	imes 10^{9} \; Nm^{2} C^{-2}$.$a_{e} = \frac{9 	imes 10^{9} 	imes (1.6 	imes 10^{-19})^{2}}{9 	imes 10^{-31} 	imes (1.6 	imes 10^{-10})^{2}}$.$a_{e} = \frac{9 	imes 10^{9} 	imes 2.56 	imes 10^{-38}}{9 	imes 10^{-31} 	imes 2.56 	imes 10^{-20}}$.$a_{e} = \frac{10^{9} 	imes 10^{-38}}{10^{-31} 	imes 10^{-20}} = \frac{10^{-29}}{10^{-51}} = 10^{22} \; m/s^{2}$.

The acceleration of an electron due to the mutual attraction between the electron and a proton when they are $1.6 \; \mathring{A}$ apart is,$(m_{e} \simeq 9 \times 10^{-31} \; kg, e = 1.6 \times 10^{-19} \; C)$. (Take $\frac{1}{4 \pi \varepsilon_{0}} = 9 \times 10^{9} \; Nm^{2} C^{-2}$)

Similar Questions

$ABC$ is a right-angled triangle in which $AB = 3 \ cm$,$BC = 4 \ cm$ and the right angle is at $B$. Three charges $+15 \ \mu C$,$+12 \ \mu C$ and $-20 \ \mu C$ are placed respectively at $A$,$B$ and $C$. The force acting on the charge at $B$ is (in $N$)

Four charges equal to $-Q$ are placed at the four corners of a square of side length $a$,and a charge $q$ is at its centre. If the system is in equilibrium,the value of $q$ is:

Three equal charges are placed on the three corners of a square. If the force between $q_1$ and $q_2$ is $F_{12}$ and that between $q_1$ and $q_3$ is $F_{13}$,the ratio of magnitudes $\frac{F_{12}}{F_{13}}$ is

$A$ point charge $A$ of $+10 \mu C$ and another point charge $B$ of $+20 \mu C$ are kept $1 \ m$ apart in free space. The electrostatic force on $A$ due to $B$ is $F_1$ and the electrostatic force on $B$ due to $A$ is $F_2$. Then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module