Capacitance of a parallel plate capacitor …
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Posted by Kunal Dahiya 8 years, 5 months ago
- 2 answers
Payal Singh 8 years, 5 months ago
Let us take a parallel plate capacitor. Suppose the separation distance between the plates is d. Use air or vacuum as a medium for this experiment.
Suppose +Q is the charge on one plate and –Q is charge on the second plate. Bring a rectangular slab made up of conducting material between the plates of the capacitor. The thickness of the slab must be less than the distance between the plates of the capacitor. When the electric field will be applied then polarization of molecules will be started. The polarization will take place in the direction same as that of electric field. Consider a vector that must be polarized, name it as P. The polarization vector must be in the direction of electric field Eo. Then this vector will start its functioning and will produce an electric field Ep in the opposite direction to that of Eo. The net electric field in the circuit is shown by the figure.
E= Eo – Ep
The electric field Eo in the outside region of the dielectric will be null. Now the equation of the potential difference between the plates will be :
V=o (d-t) + Et
But Eo= Er or K
Therefore E= Eo / k
So
V= E o (d-t) + Eot / k
V= E o [d-t+t/k]
As we know
Eo = /
= Q / A
V= Q / A [d-t+t/k]
Capacitance of the capacitor is shown in the equation below:
C= Q / V= A / (d-t+t/k)
= A /d-t (1-1/K)
I.e. C= A/ d-t (1-1/k) ——- 2.31
So, C > Co
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Naveen Sharma 8 years, 5 months ago
In Fig parallel plate capacitor inserted with dielectric slab in the gap.
When there is vacuum the the capacitance of the parallel plate capacitor is E.
Due to polarization of the dielectric the electric field is reduces to E from E . So the potential difference between the plates is,
Now if the inserted slab is conducting then electric field inside slab will be zero. Thus
So by inserting a conducting slab capacitance increases as the effective separation between the plates decrease.
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