Confusion about the charge of electrons while orbiting through bohr's energy shells
The main confusion I can see in your question is charge vs. energy. Energy is the capacity to do work. Energy comes in many forms. Kinetic energy is the energy of motion. Potential energy is the energy of the system due to its arrangement. Electrons in Bohr shells have energies dependent on the mass of the electron, which is constant, the charge of the electron (constant), the distance from the nucleus (fixed at certain allowable values), and some fundamental constants. Thus, an electron in a shell has unchanging energy. The only way for the electron to change its energy is to move to a different shell. $\Delta E=0$ for an electron in a Bohr shell; $E\neq0$.

An electron in a Bohr shell certainly has kinetic energy because it has orbital angular motion. An electron in a Bohr shell certainly has potential energy because it is being held apart from the oppositely charged nucleus. The total energy of the electron in a Bohr shell ($E_n$) is a function of the principal quantum number, $n$. $Z$ is the number of protons (1 for hydrogen, the only atom for which the Bohr model works) and $R_E$ is the Rydberg energy (related to the Rydberg constant by $R_H=R_E/hc$).

$$E_n=-\frac{Z^2 R_E}{n^2}$$

The Rydberg energy is the smash up of some other more fundamental constants, like the mass of an electron ($m_e = 9.109\times10^{-34} \text{ kg}$), the fundamental unit of charge ($e=1.602\times 10^{-19} \ \text{C}$), Coulomb's constant ($k_e = 8.987 \times 10^9 \ \ce{N m^2 C^{-2}}$ also $k_e=(4\pi\varepsilon_0)^{-1}$, where $\varepsilon_0=8.854\times 10^{-12} \ \text{F/m}$ is the permittivity of a vacuum), and the Bohr radius ($a_0 = 5.29\times 10^{-11} \ \text{m}$). The Bohr radius is itself equal to a mashup of other constants  $a_0=\hbar/(m_e c \alpha)$, where $c=2.998\times10^8 \ \text{m/s}$ is the speed of light in a vacuum, $\hbar=1.054\times10^{-34} \ \text{J s}$ is the reduced Planck constant, and $\alpha=7.297\times^{-3}$ is the fine-structure constant. Eventually, $R_E$ simplifies down to a relativistic energy:

$$R_E = \frac{k_e e^2}{2a_o}= \frac{1}{2}m_e c^2 \alpha$$

Electric charge is something else. It is a unit of electrostatic interaction with other charged particles. The electric charge of an object influences the energy of that object in the presence of an electric field, see Coulomb's Law. The (magnitude of) charge on an electron ($e$) is part of the Rydberg energy, so it influences the total energy of the electron.

However, charge$\ \neq\ $energy.