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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 / Qd / ∈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 = K∈0A / d = 4 x 8.85 x 10-12 x 1 / 2 x 10-3 = 1.77 x 10-8 farad