Question #669ef

Apr 9, 2017

An electric field $\vec{E}$ will exist between any two points due to presence of potential difference between these points.

The product of any charge $q$ and of the difference in electric potential between any two points $A \mathmr{and} B$, by definition, is the difference in electric potential energy of that charge between those points. Therefore,
${P}_{B} - {P}_{A} = q \times \left({V}_{B} - {V}_{A}\right) J$

Therefore, when a charged particle is released from rest in a uniform electric field, the corresponding electric potential at the particle's changing locations changes.

The total energy $E$ of the electron consists of its electric potential energy $P E$, and the kinetic energy $K E$. Hence,

$E = P E + K E$
where we know that if $m \mathmr{and} v$ be mass and velocity respectively of the charged particle.
$K E = \setminus \frac{1}{2} \setminus m \setminus {v}^{2}$

These are depicted in the figure below. For electron.
Initially, $K E = 0$, as it is at rest. It has $P E$ due to potential difference or due to electric field. Also due to $- v e$ charge of electron its $P E$ is $- v e$. As electron is accelerated towards $+ v e$ potential, its $K E$ increases and potential energy decreases. Clearly electric potential at the electron's changing locations must increase so that it accelerates forward.

For proton.
As in the case of electron, proton's initial $K E = 0$. Proton also has $P E$ due to potential difference or due to electric field. However, due to its $+ v e$ charge its $P E$ is $+ v e$. As positive charge accelerates towards $- v e$ potential its $K E$ increases and potential energy decreases. Since it is accelerating towards $- v e$ potential, electric potential at the proton's changing locations must decrease so that it accelerates forward.

Hope it helps.