Capacitor Part – 10
A capacitor has rectangular plates of length a and width b. The top plate is inclined at a small angle . Find the capacitance of the capacitor.
Let us consider the given capacitor is made of large number of elementary capacitors connected in parallel. Let an elementary capacitor of width dx at a distance x as shown in figure. The area of the plate of this elementary capacitor is bdx and the separation between the plates is (d + xtan
) .The capacitance of the elementary capacitor is dC = 
.
So, the net capacitance of plates is C = 
dc
= 
= 
= [In(d + xtan)
= [In(
)]
If is very small then tan = or, C = 
[In(1 + 
)]
Now log(1 + x) = x – 
x2 + —– [here x < 1]
So, C = [ – ()2] = 
[1 – 
].
2. The capacitance of a parallel plate capacitor of plate area A (length L and width b) and separation d is C. The space between the plates is filled with two dielectric medium of dielectric constants K1 and K2 as shown in figure. Find the value of C.
Let us consider an elementary part of the capacitor of width dx at a distance x as shown in figure. This part consists of two elementary capacitors with dielectric medium K1 and K2 and thickness xtan
and (d – xtan) respectively connected in series.
So, 
= 
Or, dC = 
= 
Or, C = 
Or, C = 
Or, C = 
Again tan = 
Or, C = 
[
]
Area of the plate A = Lb
So, C = 
.
Capacitance of parallel plate capacitor when dielectric slab is partly inserted:
Let us consider two parallel plates of equal plate area are separated by a distance d. A dielectric slab of thickness t dielectric constant k is partly inserted in between the plates. Now the plate capacitor is considered as two plate capacitors
(i) parallel plate capacitor of plate area A1 filled with air with capacitance C1 = 
(ii) parallel plate capacitor of plate area A2 filled with dielectric slab of thickness t with capacitance C2 = 
.
As the potential difference between the plates are same for both capacitors then they are connected in parallel.
So net capacitance of parallel plate capacitor is C = C1 + C2 = + .
Force on dielectric slab:
(i) Force required to pull the dielectric slab from parallel plate capacitor (when battery is connected with the capacitor): Let us consider the capacitor has rectangular plates of length l and width b. The length of the dielectric slab inside the plates is x. The capacitance of parallel plate capacitor is C = 
+ 
= 
[Kx + (l – x)] = [l + x(K – 1)].
The energy of the capacitor is U = 
= 
[l + x(K – 1)]V2
[as the battery is connected with the capacitor so the potential difference between the plates remain constant]
The electrostatic force Fe = external force F required to pull the slab.
So, F = 
= 
(K – 1).
(ii) Force required to pull the dielectric slab from parallel plate capacitor (when battery is disconnected from the capacitor): Let us consider the capacitor has rectangular plates of length l and width b. The length of the dielectric slab inside the plates is x. The capacitance of parallel plate capacitor is
C = + = [Kx + (l – x)] = [l + x(K – 1)].
When the battery is disconnected the charge of the capacitor (q) remain constant.
The energy of the capacitor is U = 
= 
.
The electrostatic force Fe = external force F required to pull the slab.
So, F = – = 
.
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