Physics Topics are often described using mathematical equations, making them precise and quantifiable.
What is the Formula for Charged Stored on a Capacitor?
During charging of an uncharged capacitor, electrons are removed from one plate and transferred to the other gradually. Initially, the charge of the capacitor is zero and so the potential difference between the plates is also zero. As soon as an electron is transferred from one plate to the other, an electric field builds up in the space between the capacitor plates. This field opposes further transfer. Thus as the charge accumulates on the capacitor plates, so increasingly larger amounts of work is to be done to transfer more electrons. Hence the potential difference increases due to accumulation of charges. The energy spent for doing that work, remains stored as potential energy in the electric field between the two plates of the capacitor.
Calculation: Suppose, at ans’ moment, the charge of a capacitor be q and the potential difference between the two plates is v’.
∴ The capacitance of the capacitor; C = \(\frac{q}{v}\)
Further, when some charge of amount dq is given to the capacitor, the work done against the repulsive force due to the existing charge on the capacitor plate,
dW = vdq = \(\frac{q}{C} d q\)
∴ To give Q amount of charge, the total work done
where V = final potential difference between the plates.
This work is stored as potential energy U iii the capacitor.
Energy stored in a charged parallel plate capacitor: Let us consider a charged parallel plate capacitor [Fig]. Here,
α = area of each plate,
d = separation between the plates,
k = dielectric constant of the material between the plates,
σ = surface density of charge on each plate
So, volume between the plates = αd
and amount of charge on each plate (Q) = σa.
The capacitance of this parallel plate capacitor,
C = \(\frac{\kappa \epsilon_0 \alpha}{d}\)
where, ε0 = permittivity of air or vacuum
Therefore, the energy stored in this charged capacitor,
The unit of U is joule (J). This energy is stored in the electric field between the plates of the capacitor.
[In CGS system, the expression for U is obtained by replacing ε0 by \(\frac{1}{4 \pi}\). So, U = \(\frac{2 \pi \sigma^2 \alpha d}{k}\); its unit is erg.]
Energy stored per unit volume or energy density between the plates: Energy stored per unit volume,
u = \(\frac{U}{\alpha d}\) = \(\frac{\sigma^2}{2 \kappa \epsilon_0}\)
This is called the energy density in the electric field of the
capacitor.
Now, the electric field is uniform, except at the ends, inside a parallel plate capacitor, provided the plate area is very large compared to the separation between the plates.
From Gauss’ theorem, we already know that the uniform electric field between the two plates of a parallel plate capacitor, neglect-ing end effects, is
E = \(\frac{\sigma}{K \epsilon_0}\) ; then σ = kε0E
So, the energy density between the two plates is,
u = \(\frac{\left(\kappa \epsilon_0 E\right)^2}{2 \kappa \epsilon_0}\) = \(\frac{1}{2} \kappa \epsilon_0 E^2\) ………. (1)
For vacuum or air, k = 1 . Then equation (1) becomes
u = \(\frac{1}{2} \epsilon_0 E^2\) …….. (2)
[In CGS system, the equation (1) and (2) become, due to replacement of ε0 by \(\frac{1}{4 \pi}\) , u = \(\frac{1}{8 \pi} \kappa E^2\) and u = \(\frac{1}{8 \pi} E^2\)]
In SI, the unit of u is J ᐧ m-3 [in CGS system, it is erg ᐧ cm-3] j.
Dimension of u = \(=\frac{\text { dimension of energy }}{\text { dimension of volume }}\)
= \(\frac{\mathrm{ML}^2 \mathrm{~T}^{-2}}{\mathrm{~L}^3}\) = ML-1T-2
The expressions (1) and (2) of energy density in an electric field have been derived by considering the special case of a parallel plate capacitor. However, it can be proved that these two expressions are quite general expressions—true not only for parallel plate capacitors but also for electric fields of any other type. These expressions give the energy density, i.e., energy in a unit volume around any point in an electric field of any type. The rigorous proofs of the expressions are beyond the scope of our present discussions.
Energy density around a point in an electric field:
Suppose, a point charge q be placed at a point O [Fig.]. Another point P in air, is at a distance r from O. Then the electiic field at P,
E = \(\frac{1}{4 \pi \epsilon_0} \frac{q}{r^2}\)
So, the energy density of the electric field around the point P is,
u = \(\frac{1}{2} \epsilon_0 E^2\) = \(\frac{1}{2} \epsilon_0 \frac{q^2}{16 \pi^2 \epsilon_0^2 r^4}\) = \(\frac{q^2}{32 \pi^2 \epsilon_0 r^4}\)
Then, in a small volume u around the point P, the energy stored is,
U = uv = \(\frac{q^2 v}{32 \pi^2 \epsilon_0 r^4}\)
Numerical Examples
Example 1.
The area of each plate of a parallel plate glass capacitor is 314 cm2. Its plates are separated by a distance 1 cm. What will be the radius of a sphere having a capacitance equal to that of this capacitor? [k of glass = 8]
Solution:
Capacitance of the sphere,
C = \(\frac{\kappa \alpha}{4 \pi d}\) = \(\frac{8 \times 314}{4 \times \pi \times 1}\) ≈ 200 statF
So, radius of the sphere = 200 cm.
Example 2.
The area of each plate of a parallel plate capacitor is 22 cm2 and the plates are kept separated by a paraffin paper of thickness 1 mm. Specific inductive capacity (SIC) of paraffin paper is 2. What are the capacitance of the capacitor and the surface density of charge under a potential difference of 330 V?
Solution:
Capacitance of the capacitor,
Example 3.
The distance between the two plates of a parallel plate air capacitor is d. A piece of metal of thickness \(\frac{d}{2}\) and of area equal to that of the plates Is Inserted between the plates. Compare the capacitances In the two cases.
Solution:
In the first case, capacitance of the parallel plate capacitor, C1 = \(\frac{\epsilon_0 \alpha}{d}\)
We know that intensity of electric field inside a metal is zero. So, if a piece of metal of thickness is inserted between the plates of the capacitor, the effective distance between the two plates now becomes \(\left(d-\frac{d}{2}\right)\) = \(\frac{d}{2}\). Therefore, the capacitance of the
Example 4.
The conducting plates of a parallel plate capacitor are separated by 2 cm from each other. A dielectric slab (k = 5) of thickness 1 cm is inserted between the two plates. The distance between the plates is now so changed that the capacitance of the capacitor remains the same. What will be the new distance between the plates?
Solution:
Let the present distance between the plates be d.
According to the question,
Example 5.
Surface area of each plate of a parallel plate capacitor is 50 cm2. They are separated by 2 mm in air. It is connected with a 100V power supply. Now a dielectric
(K = 5) is inserted between its two plates. What will happen
(i) If the voltage source remains connected and
(ii) if the voltage source is absent during this Insertion? [WBJEE 01]
Solution:
We know, capacitance of parallel plate capacitor is
C = \(\frac{\kappa \epsilon_0 \alpha}{d}\)
For air, the capacitance,
C1 = \(\frac{\epsilon_0 \alpha}{d}\) [ε0 = 8.854 × 10-12F ᐧ m-1]
α = 50 cm2 = 5 × 10-12F ᐧ m2,
d = 2 mm = 2 × 10-3m]
= \(\frac{8.854 \times 10^{-12} \times 5 \times 10^{-3}}{2 \times 10^{-3}}\)
= 2.21 × 10-11F
For the dielectric (k = 5), the capacitance,
C2 = \(\frac{\kappa \epsilon_0 \alpha}{d}\) = 5 × 2.21 × 10-11 = 1.11 × 10-10F
i) If a dielectric is inserted, capacitance of a parallel plate capacitor increases. Since the capacitor is still connected to the power supply, its potential will remain constant. When the intervening medium is air, charge on the capacitor,
Q0 = capacitance × potential
= 2.21 × 10-11 × 100 = 2.21 × 10-9C
When the intervening medium is the dielectric (k = 5), charge on the capacitor,
Q = 1.11 × 100 × 100 = 1.11 × 1o-8 C
∴ Change in charge of the capacitor
= Q – Q0
= 1.11 × 10-8 – 2.21 × 10-9
= 8.89 × 10-9C
Change in potential difference = 0.
ii) If the battery is removed, charge stored remains the same.
So change in charge of the capacitor = 0.
According to the question, the potential difference between the plates of the capacitor before the insertion of the dielectric = 100V.
After the insertion of the dielectric, the potential difference between the plates is,
V = \(\frac{Q_0}{C_2}\) = \(\frac{2.21 \times 10^{-9}}{1.11 \times 10^{-10}}\) = 19.91 V
So, the potential difference decreases.
Change in potential difference = 1oo – 19.91 = 80.09V
Example 6.
Each of the two square plates of a capacitor has sides of length l. The angle θ between the two plates is very small [Fig.]. If the median between the plates is air and the minimum distance between them is t, determine the capacitance of the capacitor.
Solution:
The average distance between the plates,
d = \(\frac{t+(t+l \sin \theta)}{2}\) = t + \(\frac{l}{2} \theta\)
[∵ θ is very small, sin θ ≈ θ]
∴ Capacitance of the capacitor,
Example 7.
A parallel plate air capacitor has a capacitance of 2 pF. Now, the separation between the plates is doubled, and the space is filled with wax. If the capacitance rises to 6 pF, what Is the dielectric constant of wax? [KCEr ‘o5]
Solution:
Initial separation between the plates = d; area of each plate = α.
∴ Capacitance in the 1st case,
C1 = \(\frac{\epsilon_0 \alpha}{d}\)
Final separation between the plates = 2d; dielectric constant of wax = x.
∴ Capacitance in the 2nd case,
C2 = \(\frac{\kappa \epsilon_0 \alpha}{2 d}\)
∴ \(\frac{C_1}{C_2}\) = \(\frac{2}{\kappa}\) or, k = 2 × \(\frac{C_2}{C_1}\) = 2 × \(\frac{6 \mathrm{pF}}{2 \mathrm{pF}}\) = 6
Example 8.
The area of each plate of a parallel plate capacitor is A = 600 cm2 and their separation is d = 2.0 mm. The capacitor is connected to a 200 V dc source. Find out
(i) the uniform electric field between the plates in SI unit and
(ii) the surface density of charge on any plate. Given ε0 = 8.85 × 10-12 F ᐧ m-1. [ISC ‘03]
Solution:
Here, A = 600 cm2 = 600 × 1o-4 m2 = 6 × 10-2 m2; d = 2.0 mm = 2 × 10-3m.
i) The uniform electric field between the plates,
E = \(\frac{V}{d}\) = \(\frac{200}{2 \times 10^{-3}}\) = 105V ᐧ m-1
ii) If σ = surface density of charge on any plate, then
E = \(\frac{\sigma}{\epsilon_0}\)
∴ σ = ε0E = (8.85 × 10-12) × 105
= 8.85 × 10-17 C ᐧ m-2
Example 9.
The potential of a capacitor increases from zero to 150 V when a charge of 10 esu Is Imparted to it. What will be the energy stored in the capacitor’ [HS ‘06]
Solution:
Energy stored within the capacitor
= \(\frac{1}{2}\)QV= \(\frac{1}{2}\) × 10 × \(\frac{150}{300}\) = 2.5erg
Example 10.
Find out the energy content in a volume of 1 cm3 around a point, situated In the electric field of a point charge of 10 C, at a distance of 2 m in air from the position of the point charge.
Given, ε0 = 8.85 × 10-12 F ᐧ m-1.
Solution:
The electric field at the referred point due to the point charge,
Example 11.
Estimate the percentage change of the energy stored in a parallel plate capacitor, if the separation between its plates is reduced by 10%, keeping the voltage of the charging source unchanged.
Solution:
Let d = initial separation between the plates.
∴ Final separation,
d’ = d – (10% of d) = d – \(\frac{d}{10}\) = 0.9d
Initial and final values of the energy stored in the capacitor are,
Example 12.
A 900 pF capacitor is charged to 100V by a battery. How much energy Is stored In the capacitor? (WBCHSE Sample Question)
Solution:
Here, C = 900 pF 900 × 10-12LF = 9 × 10-10 F; V = 100V
Energy stored within the capacitor,
E = \(\frac{1}{2}\)CV2 = \(\frac{1}{2}\) × (9 × 1010) × (100)2 = 4.5 × 10-6 J
Example 13.
ii The capacitance of a parallel plate air capacitor is C. The capacitor is immersed half way into an oil of dielectric constant 1.6 with the plates perpendicular to the surface of the oil. What wIll be the capacitance of this capacitor?
Solution:
The capacitance of the half of the capacitor immersed in oil,
