derive expression for capacity of a parallel plate capacitor when dielectric slab is inserted between plates
Answers
Answer:
Cm = C × ξ
Explanation:
Consider a parallel plate capacitor consisting of two plates.
Distance between the plates = d
Area of a plate = A
Charge on the a plate = Q
Potential difference between the plates = V
Capacitance of capacitor = C
If there is free space between two parallel plates of capacitor then
According to Capacitor equation
Q = C × V
⇒ C = Q / V .....(1)
We know that
V = E × d where E in electric field intensity
Putting "V = E × d " in equation (1) we get
C = Q / E × d ......(2)
According to the Gauss,s law
E = σ / ε
Putting " E = σ / ε " in equation (2) we get
C = ( Q × ε ) / ( σ × d ) ..............(3)
We know sigma is the charge density that is
σ = Q / A
Putting the "σ = Q / A" in equation (3) we get
C = A ε / d ...........(4)
This is the expression for capacitance of parallel plate capacitor without dielectric
Now
When dielectric present
Cm = ( A × ε × ξ ) / d Here Cm is capacitance in dielectric presence
Cm = ( A × ε / d ) × ξ = C × ξ ∵ C = A ε / d
Cm = C × ξ ......(5)
It shows that when we insert any dielectric plate other then the air then capacitance increase by factor ξ which is called dielectric constant.
Definition of dielectric constant
The ratio of the capacitance of a parallel plate capacitor with an insulation substance (dielectric) between the plate to the capacitance with vacuum or air as medium between the plates is called dielectric constant.
Answer:
Explanation:
The charges required to produced a certain difference of potential between the plates of a condenser is a constant ratio to the potential. This constant ratio is called capacity of condenser.
C=\cfrac{Q}{V}C=
V
Q
Consider a parallel plate capacitor having two plane metallic plate AA and BB placed parallel to each other. The plates carry equal and opposite charges +Q+Q and -Q−Q respectively.
Let AA be the area of each plate dd be the separation between the plates
\sigma=\cfrac{Q}{A}σ=
A
Q
The electric field strength between the plates
E=\cfrac{\sigma}{\epsilon_o}\longrightarrow(i)E=
ϵ
o
σ
⟶(i)
Potential difference between the plates V_{AB}=Ed=\cfrac{\sigma d}{\epsilon_o}\longrightarrow(ii)V
AB
=Ed=
ϵ
o
σd
⟶(ii)
V_{AB}=\cfrac{Qd}{\epsilon_oA}V
AB
=
ϵ
o
A
Qd
Capacitance of capacitor
C=\cfrac{Q}{V_{AB}}=\cfrac{Q}{\cfrac{Qd}{\epsilon_oA}}C=
V
AB
Q
=
ϵ
o
A
Qd
Q
C=\cfrac{\epsilon_oA}{d}C=
d
ϵ
o
A