Parallel Plate Capacitor Assignment Help

Electrostatics - Parallel Plate Capacitor

Parallel Plate Capacitor Assignment Help, Parallel Plate Capacitor Homework Help, Parallel Plate Capacitor Tutors, Parallel Plate Capacitor Solutions, Parallel Plate Capacitor Tutors, Electrostatics Help, Physics Tutors, Parallel Plate Capacitor Questions Answers

Parallel Plate Capacitor

This is the capacitor which is used most commonly. It consist of two thin conducting plates 1 and 2, each of area A held parallel to each other suitable distance d apart. The plates are separated by an insulating medium like air paper mica glass etc. One of the plated 1 is insulated and the other plate 2 is earth connected 

When a charge + Q is given to the insulated plate 1, then a charge – Q is induced on the nearer face of plate 2 and + Q is induced on the farther face of plate 2. As plate 2 is earthed the charge + Q being free, flows to earth.

Plate 1 has surface charge density σ = Q / A and plate 2 has a surface charge density – σ.

We assume that distance (d) between the plates is much smaller than the linear dimension of the plates (d2<<A). Therefore, we can use the result on electric filed due to two thin infinite parallel sheets of charge as detailed in Art.

In the regions on left of plate 1 and on right of plate 2, the electric field = 0

In the region between the plates, separated by air/ vacuum electric intensity is 

E = σ / ∈0 = 1 / ∈0 Q /A

Taking this electric field localized between the two plated as uniform throughout the potential difference V between the plates is simply the electric field times the distance between the plates,

V = E x d = 1 / ∈0 Q / A s 

The capacity C of parallel plate capacitor is given by 
C =  Q / V = Q / Q/ ∈0A = ∈0A / d
C = 
0A / d

Note. When plates of capacitor are separated by a dielectric medium of relative permittivity  ∈ r = K, is capacity 

Cm = ∈A/d = ∈r 0A / d = ∈r Co = K Co

Cm = K Co

Capacity becomes K times the capacity with air/ vacuum as dielectric.

Calculate the capacitance of a parallel plate condenser of two plates 10 cm x 10 cm each separated by 2 mm thick glass sheet of K = 4.

Sol here, A = 100 cm x 100 cm = 104 cm2 = 1m2

D = 2 mm = 2 x 10 -3m,

K = 4, C =?

C = K0A / d = 4 x 8.85 x 10-12 x 1 / 2 x 10-3 = 1.77 x 10-8 farad

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