The intensity of electric field at a point between the plates of a charged capacitor

Show that the magnetic field B at a point in between the plates of a parallel plate capacitor during charging is (mu_(0)epsilon_(0)r)/(2) (dE)/(dt) (symbols having usual meaing). ,

How does the electric field (E) between the plates of a charged cylindrical capacitor vary with the distance r from the axis of the cylinder ?

A parallel plate capacitor of plate area A and plates separation distance d is charged by applying a potential V_(0) between the plates. The dielectric constant of the medium between the plates is K. What is the uniform electric field E between the plates of the capacitor ?

Each of two large conducting parallel plates has one sided surface area A. If one of the plates is given a charge Q whereas the other is neutral, then the electric field at a point in between the plates is given by

A parallel plate capacitor has an electric field of 10^(5)V//m between the plates. If the charge on the capacitor plate is 1muC , then force on each capacitor plate is-

Two large conducting plates X and Y. each having large surface area A (on one side), are placed parallel to each other as shown in figure (30-E7). The plate X is given a charge Q whereas the other is neutral. Find (a) the surface charge density at the inner surface of the plates X, (b) the electric field at a point to the left of the plates, (c) the electric field at a point in between the plates and (d) the electric field at a point to the right of the plates.

Show that the force on each plate of a parallel plate capacitor has a magnitude equal to (½) QE , where Q is the charge on the capacitor, and E is the magnitude of electric field between the plates. Explain the origin of the factor ½ .

Figure shows a parallel plate capacitor with plate area A and plate separation d . A potential difference is being applied between the plates. The battery is then disconnected and a dielectric slab of dieletric constant K is placed in between the plates of the capacitor as shown. The electric field in the gaps between the plates and the electric slab will be

A parallel plate capacitor each with plate area A and separation ‘d’ is charged to a potential difference V. The battery used to charge it is then disconnected. A dielectric slab of thickness d and dielectric constant K is now placed between the plates. What change if any, will take place in (i) charge on the plates (ii) electric field intensity between the plates, (iii) capacitance of the capacitor Justify your answer in each case

Figue shows a parallel plate capacitor with plate area A and plate separation d . A potential difference is being applied between the plates. The battery is then disconnected and a dielectric slab of dieletric constant K is placed in between the plates of the capacitor as shown. The electric field in the dielectric slab is