The idea behind Bohr’s model is that the force required to make the electron move in a circular path is provided by the electric force of attraction between the electron and the proton. Thus, as with any circular motion, we set the force acting on the electron equal to its mass times its centripetal acceleration. This allows us to solve for the centripetal acceleration, $a_{rm cp} = v^2/r$, which in turn gives us the speed, $v$. That is,

$$
begin{align*}
kfrac{|q_1||q_2|}{r^2} &= m_{rm e}a_{rm cp}\
kfrac{e^2}{r^2} &= m_{rm e}frac{v^2}{r}\
v &= esqrt{frac{k}{m_{rm e}r}}
end{align*}
$$

The second Bohr orbit has a radius $r = 4(5.29times10^{-11};mathrm{m}) = 2.12times10^{-10}$ m, hence

$$
begin{align*}
v &= (1.60times10^{-19};mathrm{C})timessqrt{frac{8.99times10^9;mathrm{N}cdotmathrm{m}^2/mathrm{C}^2}{(9.11times10^{-31};mathrm{kg})(21.1times10^{-11};mathrm{m})}}\
&= 1.09times10^6;mathrm{m/s}
end{align*}
$$

$1.09times10^6$ m/s

We are given the electrostatic force $F$ between two known point charges $q_1$ and $q_1$ and asked about the separation distance $r$. Coulomb’s law allows us to solve for $r$:

$$
begin{align*}
kfrac{q_1q_2}{r^2} &= F\
r &= sqrt{frac{kq_1q_2}{F}}\
r &= sqrt{frac{(8.99times10^9;mathrm{N}cdotmathrm{m}^2/mathrm{C}^2)(11.2times10^{-6};mathrm{C})(29.1times10^{-6};mathrm{C})}{1.77;mathrm{N}}}\
&= 1.29;mathrm{m}.
end{align*}
$$

$1.29$ m

Since the electric force $F$ is repulsive (has a positive sign) and the given point charge $q_0$ is negative, the unknown charge $q$ must be also negative. The force $F = 1.32$ N at the separation distance $r = 1.07$ m. Substituting the known values of $F,;r$, and $|q_0|$ and solving for $|q|$, we find

$$
begin{align*}
F &= kfrac{|q_0||q|}{r^2}\
|q| &= frac{Fr^2}{k|q_0|}\
q &= frac{(1.32;mathrm{N})(1.07;mathrm{m})^2}{(8.99times10^9;mathrm{N}cdot;mathrm{m}^2/;mathrm{C}^2)(6.36times10^{-6};mathrm{C})}\
&= 26.4;mumathrm{C}
end{align*}
$$

$-26.4;mu$C