Negative energy of infalling particle in a Reissner-Nordström black hole

If we have a Reissner-Nordström black hole with Q > 0, then what charge (positive or negative) must an infalling charged particle have in order for the particle to be able to obtain negative energies at the horizon?

The energy of a charged particle with charge q, infalling into a Reissner-Nordström black hole with charge Q is given by $$\epsilon = (p-qA)\cdot \zeta_{(t)}$$. In this expression $$p = p_\mu dx^\mu$$ is the momentum of the particle, $$A = A_\mu dx^\mu = -\frac{Q}{r}dt$$ is the electromagnetic gauge field, and $$\zeta_{(t)} = \zeta^\mu_{(t)} \partial_\mu$$ is the timelike Killing vector, hence we can take $$\zeta^\mu_{(t)} = [1,0,0,0]$$, so that $$\zeta_{(t)} = \partial_t$$. Note $$dx^\mu(\partial_\nu) \equiv \frac{\partial x^\mu}{\partial x^\nu} = \delta^\mu_\nu$$, hence $$dt(\partial_t) = 1$$ and $$dr(\partial_t) = 0$$.
We will take the metric to be $$ds^2 = f(r)dt^2-f^{-1}(r)dr^2$$ and we will ignore $$\theta$$ and $$\phi$$ by assuming a purely radial trajectory for the particle as it approaches the horizon at $$r = r_h$$, where $$f(r_h) \equiv 0$$. We do not need to specify $$f(r)$$ for our problem but it is the standard one for the Reissner-Nordström black hole.
Now we evaluate $$\epsilon = (p_t - qA_t) = (p_t + \frac{qQ}{r}) = (f(r)p^t + \frac{qQ}{r})$$, where we used $$p_t = g_{tt}p^t = f(r)p^t$$. Given $$ds^2 = d\tau^2 = f(r)dt^2-f^{-1}(r)dr^2$$ and $$p^\mu \equiv \frac{dx^\mu}{d\tau}$$, we can use the first to write $$\frac{dt}{d\tau} \equiv \dot{t} \equiv p^t = (\frac{1}{f(r)}+\frac{(p^r)^2}{f^2(r)})^{1/2}$$. Substitution gives $$\epsilon = (f(r) + (p^r)^2)^{1/2} + \frac{qQ}{r}$$. Given $$\epsilon$$ is a constant of the motion we can evaluate it at $$r = r_h$$, where $$f(r_h) \equiv 0$$ and $$p^r(r_h) = 0$$ from $$d\tau^2 = f(r)dt^2-f^{-1}(r)dr^2$$.
We have $$\epsilon(r_h) = \frac{qQ}{r_h}$$, from which we see that if we want $$\epsilon < 0$$, we need $$qQ < 0$$ and so if $$Q > 0$$ then $$q < 0$$.