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Physics Topics can help us understand the behavior of the natural world around us.
Define Electrical Capacitance of a Conductor On What Factors does the Capacitance of a Conductor Depend?
If a body is heated, its temperature rises. Similarly if a conductor is positively charged, its potential increases. If bodies of different materials are heated equally, their rise in temperature is not equal because the thermal capacities of the bodies are different.
Similarly, even if different conductors are charged equally, their increase in potential may not be equal. The potential of a conductor depends not only on the amount of charge possessed by it but also on its shape, surface area, nature of the surrounding medium and presence of other conductors.
For a given conductor its potential is always proportional to its charge. If a conductor is charged with Q and as a result its potential is raised by an amount V, then
Q ∝ F or, Q = CV ………… (1)
The proportionality constant C is known as the capacitance or capacity of the conductor. Its value depends on shape, surface area, nature of the surrounding medium and presence of other conductors.
From equation (1) we have,
C = \(\frac{Q}{V}\) ……… (2)
i.e., capacitance \(=\frac{\text { amount of charge }}{\text { rise of potential }}\)
If V = 1, then C = Q
Definition: The capacitance or capacity of a conductor is defined as the charge required to raise its potential by unity.
Capacitance is a scalar quantity and is always positive.
In CGS system: In CGS system, unit of capacitance is esu of capacitance. In the above equation (2), if Q = 1 esu of charge and V = 1 esu of potential, then C = 1 esu of capacitance, i.e., if 1 esu of charge raises the potential of a conductor by 1 esu, the capacitance of the conductor is defined as 1 esu. This unit is also known as statfarad (statF).
In SI: SI unit of capacitance is farad (F). It is the practical unit of capacitance. The capacitance of a conductor is said to be 1 farad if 1 coulomb of charge is required to raise the potential of the conductor by 1 volt.
Therefore, 1F = \(\frac{1 \mathrm{C}}{1 \mathrm{~V}}\)
Since farad is a very large unit, smaller units like microfarad and picofarad are most frequently used as the unit of capacitance.
1 microfarad (µF) = 10-6 farad (F)
1 picofarad (pF) = 10-12 farad (F)
Relation between farad and esu of capacitance:
1C = 3 × 109 esu of charge
1V = \(\frac{1}{300}\) esu of potential
∴ 1F = \(\frac{1 \mathrm{C}}{1 \mathrm{~V}}\) = \(\frac{3 \times 10^9 \text { esu of charge }}{\frac{1}{300} \text { esu of potential }}\)
= 9 × 1011 esu of capacitance or statF
1µF = 10-6F = 9 × 105 esu of capacitance
1pF = 10-12F = 0.9 esu of capacitance
Dimension of capacitance:
C = \(\frac{Q}{V}\) = \(\frac{Q}{W / Q}\) = \(\frac{Q^2}{W}\)
∴ [C] = \(\frac{\mathrm{T}^2 \mathrm{I}^2}{\mathrm{ML}^2 \mathrm{~T}^{-2}}\) = M-1L-2T4I2
Numerical Examples
Example 1.
The capacitance of a spherical conductor is 1µF and the charge on it is -10C. What is its potential in air?
Solution:
We know, V = \(\frac{Q}{C}\); here C = 1µF = 10-6 F, Q = -10C
∴ V = \(\frac{-10}{10^{-6}}\) = -107V
Example 2.
The potential of a conductor having 40 esu of capaci-tance is raised by 10 esu. What is the charge on the conductor? How much charge is to be given to another conductor, having capacitance three times that of the first conductor, to raise its potential three times that of the first one?
Solution:
Charge given to the first conductor,
Q1 = C1V1 = 40 × 10 = 400 esu of charge
Capacitance of the second conductor,
C2 = 3C1 = 3 × 40 = 120 esu of capacitance
Potential rise of the second conductor,
V2 = 3 × 10 = 30 esu of potential
Charge to be given to the second conductor,
Q2 = C2V2 = 120 × 30 = 3600 esu of charge
Factors Affecting Capacitance of a Conductor
A conductor, at a potential V and having a charge Q, has a capacitance, C = \(\frac{Q}{V}\).
For constant Q, C ∝ \(\frac{1}{V}\). Hence, the factors affecting V also affects the value of C of a conductor. The value of capacitance of a conductor depends on the following factors—
- surface area and shape of the conductor,
- nature of the surrounding medium and
- presence of other conductors (especially earthed ones).
Surface area and shape of the conductor: A conductor of greater size, i.e., larger surface area has larger capaci-tance. The potential of a conductor decreases with the increase of its surface area and hence its capacitance increases.
Experiment: A thin tin sheet is suspended from a charged ebonite rod. At the bottom of the sheet, a heavy metal rod is attached. This rod keeps the sheet stretched [Fig.], The tin
sheet is connected to the disc of a gold-leaf electroscope by a metal wire. If the sheet is given a definite amount of charge, the leaves of the gold-leaf electroscope spread apart. The divergence of the leaves indicates the potential of the sheet.
Now if the tin sheet is rolled up to some extent with the help of the ebonite rod, the divergence of the leaves of the electroscope will increase. It indicates that the potential of the sheet has increased. So, charge of the sheet remaining constant, its capacitance decreases. If the area of the tin sheet is increased, the divergence of the leaves decreases, i.e., potential of the sheet decreases and its capacitance increases. So capacitance of a conductor depends on its surface area.
If the experiment is performed with conductors of the same surface area but of different shapes, it will be found that the spreading of the leaves of the gold-leaf electroscope are different. So capacitance of a conductor also depends on its shape.
Nature of the surrounding medium: if the conductor is surrounded by some dielectric medium other than air, the capacitance of the conductor increases. The effect increases with the increase of the dielectric constant of the medium.
Experiment: A charged metal sheet A placed on an insu-lating stand is connected to the disc of an uncharged gold-leaf electroscope. The divergence of the leaves of the electroscope is observed [Fig.(a)]. The amount of deflection indicates the potential of the metal sheet.
Now a dielectric slab, say a thick glass slab, is brought slowly near the sheet A. It is found that the spreading of the leaves diminishes [Fig.(b)]. So, the potential of A has decreased, i.e., its capacitance has increased. Again if the dielectric slab is removed from the vicinity of the sheet A , the leaves of the electroscope will spread out to the same extent as earlier. In this case, the dielectric medium is polarised from induction due to the metal sheet A. So opposite charges are developed on the two sides of the glass slab [Fig.(b)].
Induced negative charge reduces the potential of the sheet A and positive charge raises the potential of A . But due to the close proximity of the negative charge, its effect on the sheet A predominates. So as a whole, the potential of the sheet A diminishes a little and hence its capacitance increases.
Presence of Other conductors: The capacitance of a conductor depends on the presence of other conductors near it. If an uncharged conductor is present in the vicinity of the charged conductor under test, its capacitance increases. This effect becomes pronounced if the neighbouring conductor is earthed.
Experiment: If a positively charged conductor A placed on an insulating stand is connected to the disc of an uncharged gold-leaf electroscope, the leaves of the electroscope spread out and the amount of divergence indicates the potential of the conductor A . Now if another uncharged conductor B placed on an insulating stand is brought near A, it will be found that the divergence of the leaves diminishes a little [Fig.(a)]. So, the potential of A has decreased a little. If the conductor B is removed, the leaves spread to the same extent as earlier. From this, it is understood that if B is brought near A, the potential of A diminishes and its capacitance increases.
The reason is that an induced negative charge is developed at the nearer end and a positive charge at the far end of B due to the inducing charge of A . The induced negative charge reduces the potential of A and the induced positive charge enhances its potential. But due to the proximity of the negative charge its effect on A predominates. So as a whole, the potential of the conductor A diminishes a little and hence its capacitance increases.
Now if the conductor B is earthed [Fig.(b)], it will be found that the divergence of the leaves decreases considerably. This proves that the potential of the conductor A is highly reduced. If now the conductor B is removed from the vicinity of the con-ductor A, the leaves of the electroscope will spread out to the same extent as earlier.
So it is proved that if the conductor B is earthed, the potential of the conductor A diminishes a lot and hence its capacitance increases to a large extent. The reason is that, if B is connected to the earth, the induced positive charge being free charge moves to the earth. Under this condition, due to the presence of only negative charge in B, the potential of A diminishes a lot and hence its capacitance increases to a large extent.