Current leakage between the charged plates of a capacitor

[archaic]

Homework Statement

A charged plate capacitor is filled with a dielectric of dielectric constant $\kappa=2.50$ and resistivity $\rho=1.90\times10^{15}\,\Omega m$. It is first charged to $Q=8.00\,\mu C$.
1) What is the magnitude of the leakage current in amps?
2) What is time interval (in hours) over which the charge on the capacitor will fall to $1/e$ of its initial value?

Relevant Equations

$$I=\frac{A\Delta V}{\ell\rho}$$$$C=\frac{\kappa\epsilon_0A}{\ell}$$

1)$$\frac{Q}{\Delta V}=\frac{\kappa\epsilon_0A}{\ell}\Leftrightarrow\Delta V=\frac{Q\ell}{\kappa\epsilon_0A}$$$$I=\frac{A\Delta V}{\ell\rho}=\frac{Q}{\kappa\epsilon_0\rho}$$
2) The charge is decreasing by $\Delta Q$, so $Q(t)=Q-\Delta Q$.$$I=\frac{\Delta Q}{\Delta t}\Leftrightarrow\Delta Q=\frac{Q}{\kappa\epsilon_0\rho}\Delta t$$$$Q-\Delta Q=\frac{Q}{e}\Leftrightarrow\frac{Q}{\kappa\epsilon_0\rho}\Delta t=\frac{e-1}{e}Q\Leftrightarrow\Delta t=\left(\frac{e-1}{e}\right)\kappa\epsilon_0\rho$$
I divide by $3600$ to get the result in hours.

I am getting that 2) is wrong, though. I entered the values correctly. Here's the calculation as I have done it on desmos

Is $I=\Delta Q/\Delta t$ a wrong assumption in this case?

 

[LCSphysicist]

I think you could try take the limit in $$\Delta Q = \frac{Q}{\kappa \rho \varepsilon }\Delta t$$, not only, i think we need a negative sign here, otherwise
$$\Delta Q$$
would end being positive.

 

[archaic]

I think you could try take the limit in $$\Delta Q = \frac{Q}{\kappa \rho \varepsilon }\Delta t$$

Hm$$\frac{dQ}{dt}=-\frac{Q}{\kappa\epsilon_0\rho}\Leftrightarrow-\kappa\epsilon_0\rho\frac{dQ}{Q}=dt\Leftrightarrow-\kappa\epsilon_0\rho\ln\left(\frac{Q'}{Q_0}\right)=\Delta t$$with $Q'=Q_0/e$, I get $\Delta t=\kappa\epsilon_0\rho$.

i think we need a negative sign here

yep, I added that in $Q-\Delta Q$, but not in the first formula since the question asked for the magnitude.
This is it! Thank you. :)