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- 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?

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?