Consider the situation of the previous problem. Define displacement resistance `(R_d = V/i_d)` of the space between the plates where V is the potential difference between the plates and `i_d` is the displacement current. Show that `R_d` varies with time as ` (R_d)= (R(e^(t/tau)-1).)

A parallel plate air capacitor has capacity 'C' distance of separation between plates is 'd' and potential difference 'V' is applied between the plates force of attraction between the plates of the parallel plate air capacitor is :

Differentiate a^(x) w.r.t x, where a is a positive constant

A, B and C arc voltmeters of resistance R, 1.5R and 3R respectively as shown in the figure. When some potential difference is applied between X and Y, the voltmeter readings are V_(A) , V_(B) and V_(C) repsectively. Then :

An electron passes between two parallel plate of a capacitor as shown in the diagram. The electron enters the plate at t=0 and it comes out of the plates at time t=T. The electron hits the screen at A, as shown. Now a time varying potential difference V is applied between the plates, as shown in the graph. (Neglect the gravity). The point where the electron now will strike (as compared to point A) will be

The charge on a parallel plate capacitor varies as q=q_(0)cos 2pi v t . The plates are very large and close together (area = A, separation = d). Neglecting the edge effects, find the displacement current through th capacitor.

A slab of material of dielectric constant K has the same area as the plates of a parallel plate capacitor but has a thickless (1/2)d . Where d is the separation of the plates . How is the capacitance changed when the slab is inserted between the plates

A 20 mu F capacitor is being charged by some constant voltage source. When p.d. across capacitor changes with time at a rate 3 V/s, find conduction current in the wires joining battery with capacitor and displacement current in the space between the plates.

A parallel plate capacitor with circular plates of radius 1 m has a capacitance of 1 nF. At t = 0, it is connected for charging in series with a resistor R = 1 M Onega across a 2V battery. Calculate the magnetic field at a point P, halfway between the centre and the periphery of the plates, after t=10^(-3)s . (The charge on the capacitor at time t is q (t) = CV [1 - exp (-t//tau) ], where the time constant tau is equal to CR.)

A parallel plate capacitor with circular plates of radius 0.8 m has a capacitance of 1 nF. At t = 0, it is connected for charging in series with a resistor R = 1 M Omega across a 4V battery. Calculate the magnetic field at a point P, halfway between the centre and the perpendicular of the plates after t=10^(-3)s . (The charge on the capacitor at time t is q (t) = CV [1 - exp (-t//tau) ], where the time constant tau is equal to CR.)

A student measures the terminal potential difference (V) of a cell (of emf epsilon and internal resistance r ) as a function of the current (I) flowing through it. The slope and intercept of the graph between V and I is