you are given a `2 muF` parallel capacitor. How would you establish an instantaneous displacement current in 1mA iin the space between its plates?

Derive an expression for the capacitance of a parallel plate capacitor. How is the capacitance affected when a conducting slab is introduced in-between its plates? Explain

How would you modify 5.0 nF parallel plate capacitor storing 25.0 J of energy so that it can store 50.0 J of energy without changing the charge on its plates?

In a parallel plate capacitor with air between the plates, each has an area of 6xx10^-3m^-2 and the distance between the platesw is 3mm. Calculate the capacitance of the capacitor, If this capacitor is Connected to 100 V supply, what is the charge on each plate of the capacitor? (Given varepsilon_0=8.5xx10^-12C^2N^-1m^-2)

In these question, a statement of assertion followed by a statement of reason is given .Choose the correct answer out of the following choices: Assertion and reason both are correct statements and reason is correct explanation for assertion. Assertion and reason both are correct statements but reason is not correct explanation for assertion. Assertion is correct statement but reason is wrong statement. Assertion is wrong statement but reason is correct statement. Assertion: The displacement current goes through the gap between the plates of the capacitor ,when the charge of the capacitor does not change. Reason: The displacement current arises in the region in which the electric field and hence the electric flux does not change with the time.

A parallel capacitor has capacitance 20muF . A potential difference of 220 V is applied across it. Calculate the charge on the plates and energy stored in the capacitor. If a dieletric slab of dielectric constant 5 is introduced between its plates, calculate the new value of potential difference and the energy stored.

On charging a parallel plate capacitor to a potential V, the spacing between the plates is halved and a dielectric medium of k = 10 is introduced between the plates, without disconnectign the d.c. source. Explain, using the suitable expressions, how the capacitance

A parallel plate capacitor with air between the plates has capacitance of 9pF. The separation between its plates is 'd'. The space between the plates is now filled with two dielectrics. One of the dielectrics has dielectric constant k_1=3 and thickness d/3 while the other one has dielectric constant k_2=6 and thickness (2d)/(3) . Capacitance of the capacitor is now

On charging a parallel plate capacitor to a potential V, the spacing between the plates is halved and a dielectric medium of k = 10 is introduced between the plates, without disconnectign the d.c. source. Explain, using the suitable expressions, how the energy density of the capacity of change.

On charging a parallel plate capacitor to a potential V, the spacing between the plates is halved and a dielectric medium of k = 10 is introduced between the plates, without disconnectign the d.c. source. Explain, using the suitable expressions, how the electric field