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Advanced Physics Topics like quantum mechanics and relativity have revolutionized our understanding of the universe.
On What Factors Does the Capacitance of a Parallel Plate Capacitors Depend?
Parallel Plate Capacitor
It consists of two similar metal plates held parallel to each other, separated by a certain distance. The space in between the two plates contains air or any dielectric, e.g., glass, mica etc.
Consider two parallel plates A and B separated by a distance d. Area of each plate is ∝. Plate A is charged with a charge + Q while the plate B is grounded [Fig.]. The dielectric constant of the medium between the plates is k.
Now the surface density of charge on the plate A will be
σ = \(\frac{Q}{\alpha}\)
The inner face of the plate B is charged to -Q due to induction. If the area of the plates are large compared to the distance between them, the electric lines of force between the plates are straight and parallel except near their ends [Fig.]. Conse-quently, the intensity of electric field between the plates may be taken to be uniform. The slight deviation from this uniformity near the edges may be neglected.
The plates A and B can be considered to be infinite plates with respect to any point in between the two plates. So, the electric field at that point due to the positive charge on plate A is,
E1 = \(\frac{\sigma}{2 \kappa \epsilon_0}\), along AB [See equation (4) of section 2.5.2]
The electric field at that internal point will be in the same direc-tion AB, also due to the negative charge on plate B. Its magni-tude is
E2 = \(\frac{\sigma}{2 K \epsilon_0}\)
Therefore, at all points between the plates A and B, the electric field is
E = E1 + E2 = \(\frac{\sigma}{2 \kappa \epsilon_0}\) + \(\frac{\sigma}{2 \kappa \epsilon_0}\) = \(\frac{\sigma}{\kappa \epsilon_0}\)
If V be the potential difference between the plates, then
V = work done to bring a unit positive charge from the plate B to the plate A
= force acting on the unit charge x distance
= intensity of the electric field x distance
= E ᐧ d = \(\frac{\sigma}{\kappa \epsilon_0}\) ᐧ d
If C be the capacitance of the capacitor, then
C = \(\frac{\epsilon_0 \alpha}{d}\) = \(\frac{\sigma \alpha}{\frac{\sigma}{\kappa \epsilon_0} \times d}\) = \(\frac{\kappa \epsilon_0 \alpha}{d}\) …. (3)
[In CGS units, C = \(\frac{\kappa \alpha}{4 \pi d}\)]
For air, k = 1
Hence, C = \(\frac{\epsilon_0 \alpha}{d}\) = \(\frac{8.854 \times 10^{-12} \times \alpha}{d}\) …. (4)
Dependence of the capacitance of a parallel plate capacitor on various factors:
- Area of the plates: The capacitance is directly proportional to the area of the plates, ie., C ∝ a.
- Distance between the plates: The capacitance of a parallel plate capacitor is inversely proportional to the distance between the plates, i.e., C ∝ \(\frac{1}{d}\)
- The nature of the medium between the plates: The capacitance is directly proportional to the primittivity or dielectric constant of the medium between the plates, i.e., C K.
- The relation (3) clearly indicates that the capacitance does not depend on the charge Q or the potential V of the capacitor. Only the shape and the intervening medium determine its capacitance.
- If the common overlapping area between the two plates can be changed by using a hinge arrangement, then a changes and as a result, the capacitance C changes. This technique may be used to design a capacitor of variable capacitance.
Special case: If n number of parallel plates are alternately connected to form a multi-plate capacitor [Fig.], the capaci-tance will be,
C = \(\frac{(n-1) \kappa \epsilon_0 \alpha}{d}\)
where α = area of each plate,
d = distance between two consecutive plates
Capacitance of a parallel plate capacitor with compound dielectric: A and B are two parallel plates of which A is charged and B is grounded [Fig.]. Let α be the area of each plate, d the distance between them and + Q the charge on plate A.
So surface density of charge on each plate is σ = \(\frac{Q}{\alpha}\)
The space between the plates is now filled with two media of permittivites E1 and E2. The thickness of the two layers are (d – t) and t respectively as shown in Fig.. The intensity of the electric field between the plates may be taken as uniform.
If the dielectric constants of the two media be k1 and k2, the intensity of the electric field will be given by,
E1 = \(\frac{\sigma}{\kappa_1 \epsilon_0}\)
and E2 = \(\frac{\sigma}{\kappa_2 \epsilon_0}\); ε0 = Premittivity of air or vaccum
∴ The potential difference between the plates is,
i) Since (t – \(\frac{t}{K}\)) is a positive quantity, equation (6) shows that the capacitance of a parallel plate capacitor increases with the insertion of any dielectric medium between the plates.
ii) If a dielectric of thickness t is inserted in between the two plates of a parallel plate air capacitor, the capacitance of the capacitor becomes equal to that of a capacitor having sepa
ration between the plates reduced by (t – \(\frac{t}{K}\)), i.e., the separation between the two plates effectively decreases by the amount (t – \(\frac{t}{K}\)). If we wan to get the previous value of the capacitance, the distance between the plates is to be increased. If this increase be x, then
x = t – \(\frac{t}{K}\) = t(1 – \(\frac{1}{k}\))
iii) If n number of dielectric slabs of dielectric constants k1, k2,…. kn of thicknesses t1, t2…, tn, be inserted between the two parallel plates, the capacitance of the capacitor so formed is given by,