Is This Why My iPhone Battery Loses Its Charge So Quickly?

I don't know why, but it seems like I've solved this problem before on Brilliant. Anyways.

The situation is very similar to that of an $RC$ discharge circuit.

Hence, the charge on the capacitor decreases exponentially. Therefore,

$q = q_o e^{\frac{-t}{RC}}$

When $\displaystyle q = \frac{q_o}{2}$ ,

$\frac{q_o}{2} = q_o e^{\frac{-t_o}{RC}}$

Hence,

$t_o = RC\ln 2$

$t_o = \left(\frac{\rho l}{A}\right)\left(\frac{A\epsilon}{l}\right)\ln 2$

$t_o = \rho\epsilon\ln 2$

$t_o = 1.4\times 10^{13}\times (2.1)\times 8.85\times 10^{-12} \ln 2 = \boxed{180.4 sec}$

Voltage across the capacitor can be written as $V=IR \implies V= \rho \frac{d}{A} \frac{-dq}{dt}$ , where $d$ is the distance between the plates and $A$ is the area of each plate. ( $-dq$ since charge is decreasing)

But for capacitor, $q=CV$ at any time $t$ .

$\implies V= \frac{q}{C} = \rho \frac{d}{A} \frac{-dq}{dt} \implies dt = C \rho \frac{d}{A} \frac{-dq}{q}$

Now, put $C = \epsilon \frac{A}{d} \implies C = \epsilon_r \epsilon_o \frac{A}{d}$ , where $\epsilon_r$ is the relative permittivity. (=2.1)

$\implies dt = \rho \epsilon_r \epsilon_o \frac{-dq}{q}$

Integrating both sides $\int_0^t \,dt.= \rho \epsilon_r \epsilon_o \int_q^\frac{q}{2} \frac{-\,dq}{q}$

$\implies t= \rho \epsilon_r \epsilon_o ln2 = (1.4 \times 10^{13}) \times (2.1) \times (8.87 \times 10^{-12}) \times ln2 = 180.4s$