18.3 CAPACITANCE
18.3.1 Relation between q and V
- A capacitor is a system of two conductors separated by an insulator (Fig. 18.10). The conductors have charges, say $Q_1$ and $Q_2$, and potentials $V_1$ and $V_2$.
- Usually, in practice, the two conductors have charges $Q$ and $-Q$, with potential difference $V_{ab} = V_a – V_b$ between them. We shall consider only this kind of charge configuration of the capacitor. (Even a single conductor can be used as a capacitor by assuming the other at infinity.)
- The conductors may be so charged by connecting them to the two terminals of a battery. $Q$ is called the charge of the capacitor, though this, in fact, is the charge on one of the conductors – the total charge of the capacitor is zero.
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| Figure.18.10 A system of two conductors separated by an insulator forms a capacitor. |
- The electric field in the region between the conductors is proportional to the charge $Q$. That is, if the charge on the capacitor is, say doubled, the electric field will also be doubled at every point. (This follows from the direct proportionality between field and charge implied by Coulomb’s law and the superposition principle.)
- Suppose that a capacitor is being charged by a dc battery. While the capacitor is charging, both the charge q acquired and the p.d. V across the capacitor vary together with time. Let us measure simultaneously several values of q and the corresponding values of V while the capacitor is charging. After that we plot a graph of q against V to show how the two quantities are related.
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| Figure 18.11 Graph of charge versus potential difference for a capacitor, showing that the charge stored at any instant is proportional to the potential difference across the capacitor. |
- Figure 18.11 shows how the two quantities are related. Referring to the graph, we get charge stored in capacitor at an instant ∝ p.d. across capacitor at the same instant.
q ∝ V
18.3.2 Capacitance
1. Definition
- Now, potential difference $V$ is the work done per unit positive charge in taking a small test charge from the conductor b to a against the field. Consequently, $V_{ab} = V$ is also proportional to $Q$, and the ratio $Q/V_{ab}$ is a constant:
$C=\frac{Q}{V}$
- The constant $C$ is called the capacitance of the capacitor. $C$ is independent of $Q$ or $V$, as stated above. The capacitance $C$ depends only on the geometrical configuration (shape, size, separation) of the system of two conductors. [As we shall see later, it also depends on the nature of the insulator (dielectric) separating the two conductors.]
2. Unit
- The SI unit of capacitance is 1 farad (= 1 coulomb volt−1) or 1 F = 1 CV−1.
3. Practical Unit
- The maximum electric field that a dielectric medium can withstand without break-down (of its insulating property) is called its dielectric strength; for air it is about $3 \times 10^6 \ V.m^{−1}$. For a separation between conductors of the order of 1 cm or so, this field corresponds to a potential difference of $3 \times 10^4 \ V$ between the conductors. Thus, for a capacitor to store a large amount of charge without leaking, its capacitance should be high enough so that the potential difference and hence the electric field do not exceed the break-down limits. Put differently, there is a limit to the amount of charge that can be stored on a given capacitor without significant leaking. In practice, a farad is a very big unit; the most common units are its sub-multiples $1 \mu F = 10^{−6} \ F$, $1 nF = 10^{−9} \ F$, $1 pF = 10^{−12} \ F$, etc. Besides its use in storing charge, a capacitor is a key element of most ac circuits with important functions.
4. Symbol Representing Capacitor
- A capacitor with fixed capacitance is symbolically shown as Fig.2a, while the one with variable capacitance is shown as Fig.2b.
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| Fig.2: Capacitance is Symbolically |
Equation C = Q/V shows that for large C, V is small for a given Q. This means a capacitor with large capacitance can hold large amount of charge Q at a relatively small V. This is of practical importance. High potential difference implies strong electric field around the conductors. A strong electric field can ionise the surrounding air and accelerate the charges so produced to the oppositely charged plates, thereby neutralising the charge on the capacitor plates, at least partly. In other words, the charge of the capacitor leaks away due to the reduction in insulating power of the intervening medium.
Example 18.1
(a) How much charge is on each plate of a $4.00 \mu F$ capacitor when it is connected to a 12.0 V battery? (b) If this same capacitor is connected to a 1.50 V battery, what charge is stored?
Answer
The charge Q stored on a capacitor is given by: $Q=CV$
(a) Substitute the given values into the formula:
$Q=(4.00 \times 10^{-6} \ F)(12.0 \ V)=4.80 \times 10^{-5} \ C=48.0 \mu C$
(b) $Q=(4.00 \times 10^{-6} \ F)(1.50 \ V)=6.00 \times 10^{-6} \ C=6.00 \mu C$
Example 18.2
Two conductors having net charges of $+10.0 \mu C$ and $-10.0 \mu C$ have a potential difference of 10.0 V between them. (a) Determine the capacitance of the system. (b) What is the potential difference between the two conductors if the charges on each are increased to $+100 \mu C$ and $-100 \mu C$?
Answer
(a) Substitute the values into the capacitance formula:
$C=\frac{Q}{V}=\frac{10.0 \times 10^{-6} \ C}{10.0 \ V}$
$C=1.00 \times 10^{-6} \ F=1.00 \ \mu F$
(b) Using the formula:
$V=\frac{Q}{C}$
$V=\frac{+100.0 \times 10^{-6} \ C}{1.00 \times 10^{-6} \ F}=100.0 \ V$
So, the potential difference between the two conductors 100 V when the charges are increased to +100 μC and −100 μC.
Capacitance Examples and Graph Analysis
Example 18.3
Figure 18.12 shows how the p.d. V across a 50 µF capacitor varies with time.
(a) Sketch a graph to show how the charge q stored by the capacitor, initially uncharged, varies with time. Write down on the graph the value of q at time T.
Figure 18.12 Sketch a graph to show how the charge q stored by the capacitor, initially uncharged, varies with time
(b) The 50 µF capacitor is replaced by another one of 100 µF. The same time-varying p.d. as shown in Figure 18.7 is applied to the new capacitor. Add another line or curve to the graph drawn in (a) above to show how q varies with time for the new capacitor. Again write down the charge stored by the new capacitor at time T.
Answer
(a) At time T,
V = 7.0 V
q = CV = (50 µF)(7 V) = 350 µC
We have:
q ∝ V
Hence the shape of the q-time graph will be the same as that of the V-time graph, as shown in Figure 18.13.
Figure 18.13 The shape of the q-time graph will be the same as that of the V-time graph
(b) q = (100 µF)(7 V) = 700 µC
Example 18.4
A capacitor is composed of two conductors. One conductor carries a charge of +48 µC while the other carries a charge of -48 µC. A voltage of 12 V exists across the two conductors. Determine the capacitance of the capacitor.
Answer
Given:
Q = 48 µC = 48 × 10-6 C
V = 12 V
Using the formula:
C = Q/V
C = (48 × 10-6) / 12 = 4 × 10-6 F = 4.0 µF
Example 18.4
A capacitor is composed of two conductors. A p.d. of 12 V is applied to the capacitor. It is found that when the capacitor has become fully charged, 3.6 × 1014 electrons have moved out of one conductor and the same number of electrons have moved into the other conductor. Determine the capacitance of the capacitor.
Answer
Given:
Number of electrons n = 3.6 × 1014
Charge of an electron e = 1.6 × 10-19 C
V = 12 V
Calculate the charge:
Q = ne
Q = (3.6 × 1014)(1.6 × 10-19) = 5.76 × 10-5 C
Using:
C = Q/V
C = (5.76 × 10-5) / 12 = 4.8 × 10-6 F = 4.8 µF
Example 18.5
A variable dc voltage supply is connected to a 10 µF capacitor. The voltage increases at a constant rate from 0 to 6.0 V in the first 2 s. The voltage remains at 6.0 V for the next 2 s. After that the voltage decreases at a constant rate to zero in 2 s. Sketch a graph to show how the charge in the capacitor varies with time.
Answer
Given:
C = 10 µF = 10 × 10-6 F
Using the relation:
q = CV
Since q ∝ V, the q–time graph follows the same shape as the V–time graph, but with a different scale.
Graph stages:
0 – 2 s (V increases from 0 → 6 V)
q increases linearly from 0 to (10 µF)(6V) = 60 µC
2 – 4 s (V constant at 6 V)
q = 60 µC (constant)
4 – 6 s (V decreases from 6 → 0 V)
q decreases linearly from 60 µC to 0
Figure 18.14 Charge vs time for capacitor
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