What kind of energies are present in an electron?
From your question I assume you are talking about an electron bound to a nucleus, probably hydrogen. Since the mass of the nucleus is ~1836 times larger than the mass of the electron, we may assume the nucleus to be stationary. In a classical model of the problem, we thus have an electron with a certain velocity $v_n$ and kinetic energy $\frac{1}{2}m_ev_n^2$ at a position $r_n$ from the nucleus where it is bound by a Coulombic potential $-Ze^2/(4\pi\epsilon_0r_n)$ (where for hydrogen $Z=1$). The total energy is given by the sum of the two

$$
E_n=\frac{1}{2}m_ev_n^2-\frac{Ze^2}{4\pi\epsilon_0r_n}
$$

In order for the electron to make a (classical) stable trajectory around the nucleus, the electrostatic force should be equal but opposite to the centripedal force, hence

$$
\frac{m_ev_n^2}{r_n}=\frac{Ze^2}{4\pi\epsilon_0r_n^2}
$$ 
from which it follows that $\frac{1}{2}m_ev_n^2=Ze^2/(8\pi\epsilon_0r_n)$ and the total energy is given by
$$
E_n=\frac{Ze^2}{8\pi\epsilon_0r_n}-\frac{Ze^2}{4\pi\epsilon_0r_n}=-\frac{Ze^2}{8\pi\epsilon_0r_n}
$$
This equation shows that the total (average) kinetic energy in the hydrogen atom is half the size and opposite to the (avarage) potential energy, i.e., $\langle E_\text{kin}\rangle=-\frac{1}{2}\langle E_\text{pot}\rangle$, a result that is also predicted by the virial theorem. 

Using Bohr's postulate concerning the quantization of angular momentum, the last expression can be recast in the familiar expression for the hydrogen energies.