Leakage Current

To solve the problem I started deducing the capacitor discharge equation, then I just plugged in the values for C and R and voilà, it's done! Consider the simple RC circuit bellow: Circuit

Since the electric tension between the two components are the same we can write:

$R i + \frac{Q}{C} = 0$

But we know from the current definition that $i = \frac{dQ}{dt}$

$R \frac{dQ}{dt} = -\frac{Q}{C}$

Solving this differential equation we end up with:

$Q = Q_0 e^{-\frac{t}{RC}}$

But we want to know the time when $Q = \frac{Q_0}{2}$ :

$e^{-\frac{t}{RC}} = \frac{1}{2}$

$\frac{t}{RC} =\ln{2}$

$\boxed{t = \ln{2} R C}$

From the definition of R in terms of resistance, length, and cross section area, and from definition of C in terms of relative permittivity , length, and cross section area we have:

$R = \frac{\rho L}{A}$

$C = \frac{\varepsilon_0\epsilon A}{L}$

Solving for the time:

$t = \ln{2} \rho\varepsilon_0 \epsilon$

Using $\varepsilon_0 = 8.8542 \times 10^{-12} F/m$ :

$\boxed{ t = 180.4 s }$

Since the ceramic has a non-zero resistivity, it will act as an resistor of resistance: $R = \frac {\rho. d}{S}$ , where d is the distance between the plates and S is the surface area of the capacitor. Since the current passing on the resistor and the capacitor is the same, we can draw the circuit as an RC - circuit, and use Kirchoff's Law as usual to find the charge as a function of time on the process of discharging the capacitor. The capacitance of the capacitor is given by: $C = \frac{\epsilon.\epsilon_{0}.S}{d}$ . The product R.C will be, then: $\tau = R.C = \rho.\epsilon.\epsilon_{0}$ . The equation relating the charge on a discharging capacitor with time (on a RC - Circuit) is given by: $Q(t) = Q_{0}.e^\frac {-t}{\tau}$ . So, when $Q(t_{1}) = \frac {Q_{0}}{2}$ , we have $t_{1} = \tau.ln(2) \rightarrow t_{1} = \rho.\epsilon.\epsilon_{0}.ln(2) \approx 180.4s$ , where i've used $\epsilon_{0} = 8,854.10^{-12} F/m$ .

Let the width of the parallel palte capacitor be d, area of the plate capacitor be A, the initial charge on the capacitor be Q, and the charge at any instant on the capacitor be q. i.e. the parallel plates initially have +Q and -Q on the two sides and then at any instant they have +q and -q charge on the two sides. Let the capacitance be C. Let also the potential difference between the two plate at any instant be V, the current flowing thrhogh the plate be I, resistance of the dielectric between the two plates be R.

$I= \frac {\mathrm{d}q}{\mathrm {d} t} = \frac {V}{R}$

$V, q, C$ are related by the equation $V= \frac {q}{C}$ by definition. It can also be proven that the capacitance of the parallel plate capacitor, $C= \frac{\epsilon \epsilon_0 A}{d}$

Hence, $V= \frac{qd}{\epsilon \epsilon_0 A}$

It can also be proven that $R= \frac{\rho d}{A}$ .

Thus, $I=\frac{\mathrm{d}q}{\mathrm{d}t} = \frac {V}{R} =- \frac{q}{\epsilon \epsilon_0 \rho}$ The negative sign indicates that q is decreasing over time.

$\Rightarrow \int \frac {\mathrm {d}q}{q} = \int - \frac {\mathrm{d}t}{\epsilon \epsilon_0 \rho}$

Perform the integration, we get $\mathrm{ln}(\frac{1}{2})=- \frac{t}{\epsilon \epsilon_0 \rho}$ and sub the value of $\epsilon = 2.1, \epsilon_0= 8.854 \times 10^{-12} , \rho= 1.4 \times 10^{13}$ , we get $t= 180.4 s (4 s.f.)$

When we disconnect the source, there is still some charge $Q$ left on capacitor plates. That is, there is a potential difference (voltage) between the plates, which will force the current to flow through the dielectric. This voltage depends on the amount of charge left on capacitor, and is equal to $Q/C$ , and since we have a parallel plate capacitor: $C=\epsilon_{0}\epsilon \frac{S}{d}$ , where $S$ is the overlapping surface of plates, and $d$ is distance between them.

This voltage will cause charge leakage - a current through the dielectric, whose intensity will be: $I=V/R$ , where resistance is: $R=\rho \frac{d}{S}$ .

Summing up the above, we see that charge will leak through the dielectric, and that leakage rate depends on the amount of charge left on plates. So, after some short time $dt$ , small amount of charge $dQ$ leaks from one plate to the other, leaving less charge (by absolute value) on plates. After this, there is still some voltage between the plates, so there will be current/charge leakage again, but this time, it will be slower, since there is less voltage driving the current.

Recalling that by definition: $I=dQ/dt$ , we can describe the above process in this way: $R\ dQ/dt=-Q/C$ . (minus is here because charge is leaking-decreasing in time). This is a very common differential equation, which is found (with some modifications) everywhere from taxes, growth of bacterium colony and corporations, nuclear decay, to the fact that light in the room wont shut down immediately as we hit the switch. (and more).

In our case: $dQ/dt=-Q/\rho\epsilon_o\epsilon$ , which leads to: $dQ/Q=-dt/\rho\epsilon_o\epsilon$ . After integration this we finally have: $ln(Q/Q_{0})=-t/\rho\epsilon_o\epsilon$ , or: $Q=Q_{0}\ e^{-t/\tau}$ , where $\tau=\rho\epsilon_o\epsilon$

In the end,by putting $Q=Q_{0}/2$ , we get: $t_{1/2}=ln(2)\ \rho\epsilon_o\epsilon=180.35$

we know that for discharging equation is Q=Qmax(e^(-t/RC))

R=resistance C=capacitance

r=resistivity @=permeability of the medium=(2.1) * (8.85) * (10^-12)

R=rL/A C=@A/l

hence RC=@r

at Q=Qmax/2

t=@ r (ln(2))

using the calculator

t=180.4 s

$C=\epsilon \frac{A}{d}$ , where $\epsilon=k\epsilon_0$ , and $\epsilon_0=8.84\times 10^-12$ is the permittivity in air and k=2..1 $R=\rho \frac{d}{A}$ : $Q=Q_0 e^{\frac{-t}{RC}}$ . Setting $Q=\frac{Q_0}{2}$ , $t=-RCln0.5$ , and substituting the values for R and C, we get $t=180.1 s$ .

where q --- charge at any moment q'--- Initial charge. t --- time R--- Resistance of the circuit C--- capacitance of the circuit

  Here q=q'/2
   Therefore,  
                        (t/RC) = ln(2)

The dielectric placed in between the plates of the capacitor provides resistance and the value of that resistance is equal to

                                   R=(\rho)\times(l/A)
                          l---length of capacitor
                          A---area of capacitor plates

The capacitance is
C=(\epsilon)(\times)(\epsilon)'(\times)(A/l)

so

                                   t=(\epsilon)(\times)(\epsilon)'(\times)ln(2)

          which on simplifying gives   180.3499 sec

We know that $q=CV$ and that $\displaystyle C=\epsilon \epsilon_0 \frac{A}{d}$ . We also know that $V=IR$ (Ohm's Law) and that $\displaystyle R=\rho \frac{d}{A}$ . Lastly, we know, too, that $\displaystyle I=-\frac{\mbox{d} q}{\mbox{d} t}$ (there's a negative sign because in the context of this question there is current loss). Manipulating these expressions would allow us to arrive at the following:

$\displaystyle \mbox{d} t =-\frac{\epsilon \epsilon_0 \rho}{q} \mbox{d} q$

Now, we simply integrate both sides of the equation (note that $\hspace{1mm}$ $Q$ $\hspace{1mm}$ denotes the original charge on the capacitor at time $t=0$ and that $\hspace{1mm}$ $T$ $\hspace{1mm}$ denotes the time taken which we seek to find):

$\displaystyle \int_0^T \mbox{d}t=\int_Q^{\frac{1}{2}Q} -\frac{\epsilon \epsilon_0 \rho}{q} \mbox{d} q$

We will get the following:

$T=\epsilon \epsilon_0 \rho \ln 2 = \fbox{180.4 s}$ (3 s.f.)

Let A and d be the Area of the plates and distance between the plates of capacitor, respectively.

We know,

I= $\frac{dq}{dt}$ _ ... (1) _

and also,

q=CV , where C is capacitance and equals to $\frac{εA}{d}$ ... (2)

Since V=IR , where R is the resistance of dielectric and equals to $\frac{ρd}{A}$ ... (3)

Therefore,

Substituting V in (3) from (2), and then substituting I in (1),

we get,

$\frac{dq}{q}$ = $\frac{q}{CR}$ dt

=> $\frac{dq}{q}$ = $\frac{q}{ερ}$ dt

Integrating both sides

we get,

t=ρεln2

Putting the values of ρ and ε,

t= $1.4 \times 10^{13}$ X 2.1 X $8.853 \times 10^{-12}$ X 0.693

t= $\boxed{180.393}$