An electron ($mass = 9.1 \times {10^{ - 31}}kg$) is sent into an electric field of intensity $9.1 \times {10^6}$ Newton/coulomb. The acceleration produced is
A. $1.6 \times {10^{18}}m/{s^2}$
B. $1.6 \times {10^6}m/{s^2}$
C. $1.6 \times {10^{ - 18}}m/{s^2}$
D. $1.6 \times {10^{ - 6}}m/{s^2}$
Answer
599.4k+ views
Hint: The force acting on an electron is given by $F = q \times E$, where q is the charge of an electron and E is the electric field. Substitute the values given and find the force. Force is also defined as the product of mass and acceleration. Substitute the values of mass and force to find the value of acceleration.
Complete step by step answer:
We are given that an electron ($mass = 9.1 \times {10^{ - 31}}kg$) is sent into an electric field of intensity $9.1 \times {10^6}$ Newton/coulomb.
We have to find the acceleration produced by the electron.
Force acting on an electron is defined as the product of the charge of the electron and electric field it got sent to.
$F = q \times E$
Charge of an electron is $1.6 \times {10^{ - 19}}$ coulombs.
Electric field is $9.1 \times {10^6}$ Newton/coulomb.
$
F = q \times E \\
\Rightarrow F = 1.6 \times {10^{ - 19}} \times 9.1 \times {10^6} \\
\Rightarrow F = 14.56 \times {10^{ - 13}}N \\
$
We also know that force is the product of the mass of the body and acceleration it has produced.
$F = m \times a$
m is the mass of the electron and a is the acceleration
$mass = 9.1 \times {10^{ - 31}}kg$
$
F = m \times a \\
a = \dfrac{F}{m} \\
F = 14.56 \times {10^{ - 13}}N \\
\Rightarrow a = \dfrac{{14.56 \times {{10}^{ - 13}}}}{{9.1 \times {{10}^{ - 31}}}} \\
\Rightarrow a = 1.6 \times {10^{18}}m/{s^2} \\
$
The acceleration produced by the electron of $mass = 9.1 \times {10^{ - 31}}kg$ which is sent into an electric field of intensity $9.1 \times {10^6}$ Newton/coulomb is $1.6 \times {10^{18}}m/{s^2}$
The correct option is Option A, $1.6 \times {10^{18}}m/{s^2}$
Note:Electrons are the negatively charged particles of atoms. Electrons are accelerated by constantly applying an electric field that comes from the external potential difference between two points, but are decelerated by the intense internal electric fields from the material atoms that make up the circuit. An electron will move in the opposite direction of the electric field because of its negative charge.
Complete step by step answer:
We are given that an electron ($mass = 9.1 \times {10^{ - 31}}kg$) is sent into an electric field of intensity $9.1 \times {10^6}$ Newton/coulomb.
We have to find the acceleration produced by the electron.
Force acting on an electron is defined as the product of the charge of the electron and electric field it got sent to.
$F = q \times E$
Charge of an electron is $1.6 \times {10^{ - 19}}$ coulombs.
Electric field is $9.1 \times {10^6}$ Newton/coulomb.
$
F = q \times E \\
\Rightarrow F = 1.6 \times {10^{ - 19}} \times 9.1 \times {10^6} \\
\Rightarrow F = 14.56 \times {10^{ - 13}}N \\
$
We also know that force is the product of the mass of the body and acceleration it has produced.
$F = m \times a$
m is the mass of the electron and a is the acceleration
$mass = 9.1 \times {10^{ - 31}}kg$
$
F = m \times a \\
a = \dfrac{F}{m} \\
F = 14.56 \times {10^{ - 13}}N \\
\Rightarrow a = \dfrac{{14.56 \times {{10}^{ - 13}}}}{{9.1 \times {{10}^{ - 31}}}} \\
\Rightarrow a = 1.6 \times {10^{18}}m/{s^2} \\
$
The acceleration produced by the electron of $mass = 9.1 \times {10^{ - 31}}kg$ which is sent into an electric field of intensity $9.1 \times {10^6}$ Newton/coulomb is $1.6 \times {10^{18}}m/{s^2}$
The correct option is Option A, $1.6 \times {10^{18}}m/{s^2}$
Note:Electrons are the negatively charged particles of atoms. Electrons are accelerated by constantly applying an electric field that comes from the external potential difference between two points, but are decelerated by the intense internal electric fields from the material atoms that make up the circuit. An electron will move in the opposite direction of the electric field because of its negative charge.
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