How dielectric can increase the capacitance of a capacitor?

Consider a parallel-plate capacitor   ·       Initially the plates are separated by a vacuum and connected to a battery, giving the charge on the plates +Q and –Q ·       The battery is now removed and the charge on the plates remains constant ·       The electric field between the plates is uniform and has a magnitude of E0 ·       Meanwhile the separation between plates is d

·       When a dielectric is placed in the electric field between the plates, the molecules of the dielectric tend to become oriented with their positive ends pointing toward the negatively charged plate and vice versa ·       The result is a buildup of positive charge on one surface of the dielectric and of negative charge on the other

·       The number of field lines within the dielectric is reduced thus the applied electric field E0 is partially canceled ·       The new electric field, E is electric field between the plates is uniform and has a magnitude of E0 ·       The new electric field strength (E < E0) is less, then the potential difference, V across the plates is less V = Ed ·       Since V is smaller while Q remains the same the capacitance C = Q/V        is increased by the dielectric

Charging & discharging of capacitor

As soon as the switch is closed, the battery is connected across the capacitor, current flows and the potential difference across the capacitor begins to rise but, as more and more charge builds up on the capacitor plates, the current and the rate of rise of potential difference both fall. Finally no further current will flow when the p.d. across the capacitor equals that of the supply voltage V_o. 
The capacitor is then fully charged. 

Current starts to flow and the potential difference across the capacitor drops. As charge flows from one plate to the other through the resistor the charge is neutralised and so the current falls and the rate of decrease of potential difference also falls. Eventually the charge on the plates is zero and the current and potential difference are also zero - the capacitor is fully discharged.

Vbattery = Vresistor + Vcapacitor

Vbattery = IR + Q/C

Capacitor in series & parallel

CASE4: A stationary particle of charge q and mass m is placed in a uniform electric field, E

CASE3: An electron (e) with mass, me enters a uniform electric field, E perpendicularly with an initial velocity u

Gravitational potential (V), work (W) & potential energy (U)

ANSWER TECHNIQUE 1

IMPORTANT

VECTOR QUANTITY	SCALAR QUANTITY
Electrostatic force, F – Coulomb’s law	Electric potential, V
Electric field, E	Work, W
	Potential energy, U
In calculation, NO need to put the sign of charge	In calculation, NEED to put the sign of charge either (+)ve or (–)ve

Inverse square law

The inverse-square law, is quite common in Physics. This law state that for a specified physical quantity or intensity is inversely proportional to the square of the distance from the source of that physical quantity.
The fundamental cause for this can be understood as geometric dilution corresponding to point-source radiation into three-dimensional space.

Coulomb's law is one of it. Applications of the inverse square law: gravity, sound, light and radiation.

Coulomb's Law & Electric field

Google for Hyperphysics, for further explanation