I think some of your charges are a bit off for what you're trying to do. The idea (as I understand it) is that the stuff on the electric planet is attracted to the planet itself in the same way normal stuff is attracted to a normal planet.
Effects on Alice
The effects on Alice will depend on a few things. First, how is the charge distributed through her? Is every molecule in her body somehow slightly charged? If so, that's probably going to have a lot of biological effects I'm not qualified to answer. If not, the force won't be distributed perfectly across her body, so it could cause the force to be perceived differently than gravity would.
The big thing here is her inner ear. If the fluid that helps us sense acceleration has enough charge difference that it behaves properly, it would probably keep her normal sense of balance intact.
But you'd also need to account for different ways we move muscles. If all the charge is in her head, her neck muscles will get a lot stronger, while her arm muscles will be barely used. So the charge would ideally be distributed fairly evenly through her body. Still, you could probably get away with, say, a thousand pockets of charge implanted inside insulators (so the charge doesn't all flow into her body), with the pockets attached to various parts of her skeleton.
Obviously, there are some medical issues at play, but I don't think it's wholly implausible.
Setting up an Electric Planet
Warning: Lots of Math! I probably screwed something up, so don't take it as gospel truth.
Coulomb's law of electrostatic interaction works the same as Newton's law of universal gravitation. Subscript p is for planet, s is for stuff on the planet, g is for gravity, e is for electric.
$F_g={Gm_pm_s\over r^2}$ vs $F_e={kq_pq_s\over r^2}$
Now, one major difference is when we apply it to acceleration.
For both cases,
$F=MA$$\leftrightarrow A={F\over M}$
For gravity, the inertial mass is identical to the gravity-causing mass, so we can reduce the equation,
$A_g={{Gm_pm_s\over r^2}\over m_s}$$={Gm_p\over r^2}$
However, for electric charges, inertial mass has nothing to do with electric charge, so no such reduction is inherently possible,
$A_e={{kq_pq_s\over r^2}\over m_s}$$={kq_pq_s\over m_sr^2}$
This is a good thing though, or it would be impossible to have lots of charge without lots of gravity. For the next part, we want the electric force to be the same as the gravitational force on Earth. To make it feel right, this means we need to assign a value of electric charge that's relatively proportional to its mass (so small things have small charge, big things have big charge).
So, $F_g=F_e$
${Gm_pm_s\over r^2}={kq_pq_s\over r^2}$
$Gm_pm_s=kq_pq_s$
From here, we can pick some target mass, say 100 kg, then plug it in. We can look up the following constants and values:
$G$$=6.674\cdot 10^{−11} {N\cdot m^2\over kg^2}$
$k$$=8.987552\cdot 10^9 {N\cdot m^2\over C^2}$
$m_p=$$m_{earth}$$=5.9721986\cdot 10^{24} kg$
$m_s=m_{target}=100 kg$
We can plug them into the equality above to get,
$6.674\cdot 10^{−11} {N\cdot m^2\over kg^2}\cdot 5.9721986\cdot 10^{24} kg\cdot 100 kg$$=8.987552\cdot 10^9 {N\cdot m^2\over C^2}\cdot q_pq_s$
$3.986\cdot 10^{16} N\cdot m^2$$=8.987552\cdot 10^9 {N\cdot m^2\over C^2}\cdot q_pq_s$
${3.986\cdot 10^{16}\over 8.987552\cdot 10^9}C^2=q_pq_s$
$q_pq_s=4.435\cdot 10^6 C^2$
At this point, we can see that there are infinitely many solutions. Our planet can have very little charge while the stuff on it has a lot of charge, but this will cause the stuff to repel each other very much. Or the planet can have a lot of charge while the stuff on it has very little. The second idea seems better to me. The most natural way in my mind is to make the proportions of charges the same as the proportions of mass.
So, $m_p\over m_s$$={q_p\over q_s}$. For our 100 kg test case, this means we have:
$5.9721986\cdot 10^{24} kg \over 100 kg$$={q_p\over q_s}$$=5.9721986\cdot 10^{22}$
We can manipulate an earlier answer to get,
$q_pq_s=4.435\cdot 10^6 C^2$
${q_pq_s\over q_s^2}$$={4.435\cdot 10^6 C^2\over q_s^2}$
${q_p\over q_s}$$={4.435\cdot 10^6 C^2\over q_s^2}$$=5.9721986\cdot 10^{22}$
$q_s^2$$={4.435\cdot 10^6 C^2\over 5.9721986\cdot 10^{22}}$
$q_s^2=7.426\cdot 10^{-17} C^2$
$q_s=8.617\cdot 10^{-9} C$
$q_p\cdot 8.617\cdot 10^{-9} C=4.435\cdot 10^6 C^2$
$q_p={4.435\cdot 10^6 C^2\over 8.617\cdot 10^{-9} C}$
$q_p=5.1468\cdot 10^{14} C$
To set this up, we need a relatively massive object (so Alice can't noticeably move the planet every time she jumps) that is light enough to have negligible gravity still. Something like 1 megaton$=10^9 kg$ ought to do. The planet will barely accelerate when she moves since it's 10 million times her mass, and only exerts about $0.07 N$ of force which equates to about $0.0006 {m\over s^2}$ of acceleration even if the center of mass is just 10 meters under her feet. Note that this makes the planet much more dense than Earth, $238732 {kg\over m^3}$ vs. $5495 {kg\over m^3}$, and making it less dense will just decrease the gravitational effect even further.
Next, we need a highly-charged something sitting about 6400 km from Alice.
Charge per proton/electron is $\pm 1.6021766\cdot 10^{-19} C$, and our total charge was $5.1468\cdot 10^{14} C$.
This gives us,
${5.1468\cdot 10^{14} C \over 1.6021766\cdot 10^{-19} {C\over e^-}}$$=3.212\cdot 10^{33} e^-$
Obviously, we can't just have $10^{33}$ electrons hanging out somewhere, or they'll blow apart. What we really need is $10^{33}$ extra electrons in a relatively balanced system. For example, we could have $10^{38}$ protons and $10^{38}+10^{33}$ electrons. From Alice's perspective, there's no difference (well, except the part where she might actually survive), but there's a huge difference from the charged object's perspective.
One proton mass $=1.6726\cdot 10^{-27} kg$, so $10^{38}$ protons has a mass of $1.6726\cdot 10^{11} kg$. The total electron mass is only about $10^8 kg$, which is pretty negligible in comparison. Because all this mass is 6400 km from Alice, it will have negligible gravitational influence on her.
Of note, we could also do this by having a much lower charge that's much closer to Alice, but the closer we get the charge to Alice, the more important tidal effects will be. If we put the charge 10 meters under her feet, her head would feel noticeably less pull than her feet (probably -- I didn't actually do the math).
So let's solve all the above for a distance of 10 km. This lets us put Alice on a small asteroid 10 km in radius, then tunnel down to make the charge at the core. It's a lot more feasible than the construction I made at first.
We'll leave the charges for the stuff the same, which gives us,
${Gm_pm_s\over r_{Earth}^2}$$={kq_pq_s\over r_{10 km}^2}$
$q_p=Gm_pm_sq_s{r_{10 km}^2\over r_{Earth}^2}$
Noting that our previous value of $q_p$$=Gm_pm_sq_s$, we can just multiply by the ratio of squared distances,
$q_{p2}=q_{p1}{r_{10 km}^2\over r_{Earth}^2}$
$q_{p2}=5.1468\cdot 10^{14} C\cdot {(10 km)^2\over (6367 km)^2}$
$q_{p2}=1.2696\cdot 10^9 C$
This equates to $7.9\cdot 10^{27}$ electrons. Using the same $10^5:1$ ratio as before, the protons would mass about $13.3 kg$. Again, 13 kg at 10 km is going to have a negligible gravitational effect on Alice.
Note: My gut tells me that if 13 kg of charged particles can accelerate Alice at 1g from 10 km away, I screwed the math up somewhere, or this system isn't physically stable (like, the amount of charge inside Alice would boil her while it escaped into space or something). Still, the concept is valid, and I think it's plausible something like this could be done, especially for non-biological Alice-s. It seems plausible that a few hundred tons of charged stuff in the core should be physically capable of reasonably Earth-like acceleration from 10 km away.
