Parallel plate condenser
“Parallel plate condenser” :
It is the device by which we can increase the capacitance of conducting plate without changing its size.
In the above figure shows the construction of parallel plate condenser, for this we take two conducting plates A and B of equal area of cross section A. They kept parallel to each other at a distance d. when +Q charge is given to plate A, this charge uniformly distributed on outer surface and due to electrical induction negative charge -Q is produced on inner surface of plate B and outer surface of plate B is earthed. the electric field intensity at the corner of two plate in non uniform which indicates by the curve field lines but in whole length of the two plate the electric field intensity is uniform.
“Derivation of expression of capacitance”: we know that the electric field between two parallel plates of surface charge density +σ and -σ is
E = σ/εoK N/C
where, K = dielectric constant of a medium present in between two plates .
we know that by the relationship between E and ∆V
E = ∆V/∆X
Hence, ∆V = E × ∆X
(Putting the values of E)
∆V = σ/εoK × d
(∵ charge density σ = Q/A )
∆V = (Q/A·εo K) × d
By the theory of condenser
capacitance = charge on either plate /
potential diff bw two plates
C = Q/∆V
putting the value of ∆V
C = (Q×AεoK)/Q×d
C = AεoK/d farad
This is the required expression of parallel plate condenser containing medium.
#) factor affecting the capacitor
1) Area of cross section of plates:
The capacitance of parallel plate condenser is directly proportional to the area of cross section of plate
i.e C ∝ A
2) Medium present between plates:
The capacitance of parallel plate condenser directly proportional to the dielectric constant of a medium present between two plates
i.e C ∝ K
3) Distance between plates:
The capacitance of parallel plate condenser inversely proportional to the distance between plates
i.e C ∝ 1/d
It is the device by which we can increase the capacitance of conducting plate without changing its size.
In the above figure shows the construction of parallel plate condenser, for this we take two conducting plates A and B of equal area of cross section A. They kept parallel to each other at a distance d. when +Q charge is given to plate A, this charge uniformly distributed on outer surface and due to electrical induction negative charge -Q is produced on inner surface of plate B and outer surface of plate B is earthed. the electric field intensity at the corner of two plate in non uniform which indicates by the curve field lines but in whole length of the two plate the electric field intensity is uniform.
“Derivation of expression of capacitance”: we know that the electric field between two parallel plates of surface charge density +σ and -σ is
E = σ/εoK N/C
where, K = dielectric constant of a medium present in between two plates .
we know that by the relationship between E and ∆V
E = ∆V/∆X
Hence, ∆V = E × ∆X
(Putting the values of E)
∆V = σ/εoK × d
(∵ charge density σ = Q/A )
∆V = (Q/A·εo K) × d
By the theory of condenser
capacitance = charge on either plate /
potential diff bw two plates
C = Q/∆V
putting the value of ∆V
C = (Q×AεoK)/Q×d
C = AεoK/d farad
This is the required expression of parallel plate condenser containing medium.
#) factor affecting the capacitor
1) Area of cross section of plates:
The capacitance of parallel plate condenser is directly proportional to the area of cross section of plate
i.e C ∝ A
2) Medium present between plates:
The capacitance of parallel plate condenser directly proportional to the dielectric constant of a medium present between two plates
i.e C ∝ K
3) Distance between plates:
The capacitance of parallel plate condenser inversely proportional to the distance between plates
i.e C ∝ 1/d
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