16.3 Electric Potential Energy

16.3 Electric Potential Energy

We have learnt in the previous section that a charge placed in an electric field possesses some electrical potential energy. If a charge $q_{o}$ is placed at any point in the electric field due to a charge $q$ and allowed to move it freely, it will be repelled away by the charge $q$. Equation (16.17) shows that the potential decreases with the increase of distance. Therefore, if a positive charge is allowed to move freely in an electric field, it will always accelerate from a point of higher potential to a point of lower potential. If the potential difference between these points be $\Delta V$, the potential energy of the charge will decrease by $\Delta U = q_{o} \Delta V$. This decrease in potential energy of the charge will appear as its kinetic energy.

If $m$ is the mass of the body carrying charge $q_{o}$, then the body will reach the point at lower potential energy with velocity $v$, then

A negative charge allowed to move freely in an electric field moves from a point of lower potential to a point of higher potential.

The equation (16.18) suggests that the energy can also be expressed as the product of potential difference and charge. In atomic physics, we use one such unit of energy known as electron-volt (eV). This unit has already been defined in the previous class. It is the energy required by an electron which has been accelerated through a potential difference of $1$ volt.

Hence, $1\,\mathrm{eV} = \text{charge on an electron} \times 1\,\mathrm{volt}$

$1\,\mathrm{eV} = 1.6 \times 10^{-19}\,\mathrm{J}$