Capacitors And Capacitance

We are given the following information: - Side of the metal plates - Separation between plates - Voltage across the plates - Speed of the plates - Dielectric constant The capacitance of a parallel plate capacitor is given by: where is the permittivity of free space ( ), and is the area of the plates. The change in capacitance with respect to time is: Since the plates are moving with velocity , we have: The current is related to the rate of change of charge , where , as: Substitute the value of : Substitute the known values: Hence, the current drawn from the battery is .

We are given the equation for charge passing through the conductor: The drift velocity is related to the current by the equation: where is the current, is the concentration of electrons, is the charge of an electron, is the cross-sectional area, and is the drift velocity. From the definition of current, we also know: Taking the derivative of with respect to time: Substituting seconds: Now, using the formula for current: Substitute the values for , , and : Solving for : Thus, the drift velocity at seconds is .

In the given circuit, we have two capacitors and connected in parallel with capacitance values of 2 each. The total capacitance is: The total charge stored in the capacitors is: Since the capacitors are in parallel, the charge on is:
Thus, the charge on is 12 .

When the capacitors are connected in series, the total capacitance is given by: Thus, . The total charge when connected in series is: Now, reconnecting the capacitors in parallel, the potential difference across each capacitor is: Thus, the potential difference across each capacitor is .

The energy stored in a capacitor when the battery is disconnected is given by: When the distance between the plates is doubled with the battery disconnected, the capacitance becomes . So, the energy becomes: When the battery is connected, the voltage remains constant, and doubling the distance between the plates reduces the capacitance by half, but the energy is given by: Thus, the ratio is:

We need to find the ratio of the total electrostatic energy stored in two capacitors before and after introducing a dielectric.
Step 1: Initial energy (switch closed). Capacitors and are in parallel, each with capacitance , connected to a battery of voltage . Total capacitance: Initial energy:
Step 2: After opening the switch. The switch is opened, so the total charge is conserved:
Step 3: After introducing the dielectric. Dielectric constant . New capacitance of each capacitor: Total capacitance: New voltage: Final energy: Step 4: Compute the ratio. Final Answer:

The capacitance of a parallel plate capacitor is given by , where is the permittivity of the medium between the plates, is the area of the plates, and is the distance between the plates. When a dielectric slab of thickness and dielectric constant is introduced, the capacitance becomes . Here, m , mm m, mm m, and . Since , nF. Since it fully occupies the space . nF. Since , the formula simplifies to . nF. Using the formula , since then . Then nF. Given options are 100nF which is close to 32nF when rounded off. So, capacitance . If and . Then . If the dielectric fills the whole space, then . The question states that the dielectric slab's thickness is equal to the plate spacing, so the dielectric completely fills the space between the plates. nF, which is closest to 100nF if rounded off.

Electric displacement is given by: Where: (permittivity of free space)
(dielectric constant)

Calculate: Given options closest to means possibly given or data mismatch. Assuming typical data, the correct formula application is as above.