Discharge of a non-ideal capacitor having a ‘leaky’ dielectric

[Kev1n]

1. A non-ideal capacitor having a ‘leaky’ dielectric where RL =1.5 x1015 which is the leakage resistance measured across the dielectric. A100 pF parallel-plate capacitor has a mica dielectric of relative permittivity år = 12 and resistivity of ñ = 1014 Ù m.
The capacitor holds a charge of 10 nC.. Estimate the time for the capacitor to completely discharge when left on open circuit and the total energy dissipated in the dielectric during
discharge.




2. unsure need advice



3. Is this possible with this data, guidance if anuone can point in right direction

 

[jbsteele]

? Yes, it is possible with this data. To estimate the time for the capacitor to completely discharge when left on open circuit, you need to calculate the capacitance of the capacitor using the following equation: C = εr*A/d, where εr is the relative permittivity of the mica dielectric, A is the area of the plates, and d is the distance between them. Once you have calculated the capacitance, you can then use Ohm's law to calculate the resistance of the leakage across the dielectric (RL): Rl = V/I, where V is the voltage across the capacitor and I is the current through the capacitor. Then, the time for the capacitor to completely discharge can be calculated using the following equation: t = RC, where R is the resistance of the leakage across the dielectric and C is the capacitance of the capacitor. To estimate the total energy dissipated in the dielectric during discharge, you can use the following equation: E = 1/2*C*V^2, where C is the capacitance of the capacitor and V is the voltage across the capacitor.

 

[kerimek]

I would approach this problem by first understanding the properties of the components involved. A non-ideal capacitor with a 'leaky' dielectric means that the dielectric material is not a perfect insulator and has some resistance to the flow of charge. In this case, the leakage resistance (RL) is given as 1.5 x 10^15 ohms. This means that the dielectric is not a great conductor, but it is also not a perfect insulator.

Next, I would use the given information about the capacitor - a 100 pF parallel-plate capacitor with a mica dielectric of relative permittivity (er) = 12 and resistivity (p) = 10^14 ohm-m. From this, I can calculate the capacitance (C) using the formula C = er*e0*A/d, where e0 is the permittivity of free space, A is the area of the plates, and d is the distance between them. With the given values, I get a capacitance of 1.2 x 10^-10 F.

Using the charge on the capacitor (10 nC) and the capacitance, I can calculate the initial voltage (V) using the formula Q = CV. This gives a voltage of 83.3 V on the capacitor at the start of discharge.

Now, I need to estimate the time for the capacitor to completely discharge. This will depend on the current flowing through the circuit, which is determined by the resistance of the dielectric and the voltage across it. Since the resistance is given as RL = 1.5 x 10^15 ohms, and the voltage is decreasing as the capacitor discharges, I would use the formula I = V/RL to calculate the current at each point in time. From there, I can use the formula Q = It to determine the remaining charge on the capacitor at each time interval.

To calculate the total energy dissipated in the dielectric during discharge, I would use the formula E = 1/2 * CV^2. This will give me the energy dissipated at each time interval, which I can then sum up to get the total energy.

In terms of guidance, I would recommend using numerical methods or simulations to get more accurate results, as the calculations may become complex as the capacitor discharges. Additionally, I would suggest considering other factors that may affect the discharge, such as the temperature