Question

An electron of mass m and charge e is accelerated from rest through a potential difference v in vacuum. Its final speed will be?
$\eqalign{
  & {\text{A)}}\dfrac{{{\text{eV}}}}{{{\text{2m}}}} \cr
  & {\text{B)}}\dfrac{{{\text{eV}}}}{{{\text{2m}}}} \cr
  & {\text{C)}}\sqrt {\dfrac{{{\text{2eV}}}}{{\text{m}}}} \cr
  & {\text{D)}}\sqrt {\dfrac{{{\text{eV}}}}{{{\text{2m}}}}} \cr} $

Answer 1

Hint:Here in this question coulomb force will be acting on charge which is a conservative force, such that work done charge is equal to kinetic energy gained by charge.

Complete Step by step Solution:
Given that an electron is of mass m, charge e, and potential difference is v
$\eqalign{
  & {\text{Work done, by a charge in motion}} \cr
  & {\text{W = e V}} \cr
  & {\text{This work done W }}{\text{is converted into Kinetic energy}} \cr
  & \Rightarrow {\text{ Kinetic energy = W}} \cr
  & \Rightarrow \dfrac{1}{2}{\text{ m (}}{{\text{v}}^2}{\text{ - }}{{\text{u}}^2}{\text{) = eV}} \cr
  & {\text{Also u is initial speed, so u = 0}} \cr
  & \Rightarrow \dfrac{1}{2}{\text{ m }}{{\text{v}}^2}{\text{ = e V}} \cr
  & \Rightarrow {\text{ }}{{\text{v}}^{^2}} = {\text{ }}\dfrac{{2{\text{ eV}}}}{{\text{m}}} \cr
  & \therefore {\text{ v = }}\sqrt {\dfrac{{2{\text{ eV}}}}{{\text{m}}}} \cr} $

Additional Information: When a free positive charge q is accelerated by an electric field, it is given kinetic energy. The process is analogous to an object accelerated by a gravitational field. It is as if the charge is descending an electric hill where its electric potential energy is converted into kinetic energy. Let's explore the work done on a charge q by the electric field in this process, so that we can develop a definition of electric potential energy.
The electrostatic or Coulomb force is conservative, which means that the work done on q is independent of the path taken. This is exactly analogous to the gravitational force in the absence of dissipative forces such as friction. When a force is conservative, it is possible to define a potential energy associated with the force, and it is generally easier to deal with potential energy (because it depends only on position) than to calculate work directly.

Notes: The magnetic force acts in such a way that the direction of the magnetic force and the velocity are always perpendicular to each other. If the force and velocity are perpendicular, the force and displacement are also perpendicular, then W = FS cos q, if q = 90, the work done will be zero.