Page 33 - Physics_Form_2

Static electricity

their total capacitance, C, can be obtained
,
If V V 2 , and V are potential differences using equation (3) as follows:
1
3
are developed between the various plates,
then the total potential difference, V, 1 1 1
across AD is: C = C 1 + C 2

FOR ONLINE READING ONLY
V = V + 1 V + 2 V (1) CC
3
C = 1 2
1
When capacitors are in series, there is an C + C 2
equal distribution of charge on the plates.
Charge Q on C is transferred to C and Example 1.3
2
1
C by induction. Then, Three capacitors, labeled A, B, and C,
3
Q Q Q have capacitances of 10 μF, 20 μF, and
V = C , V = C , and V = C 30 μF, respectively, and are connected
2
1
3
2
1
3
in series. Find the value of a single
Substituting ,V V 2 , and V into equation capacitor that could replace them.
1
3
(1), you get;
Solution
V = Q + Q + Q (2) Use the formula for capacitors in series.
C 1 C 2 C 3 1 1 1 1
 1 1 1  C = C + C + C
V  Q     but, V = Q 1 2 3
 C 1 C 2 C 3  C 1 1 1 1
  
C 10 μF 20 μF 30 μF
Where C is the combined or equivalent
capacitance. 632 11
 
Q  1 1 1  60 μF 60 μF
Then,  Q     ,
C  C 1 C 2 C 3  60 μF
C   5.45 μF

Simplifies to: 11
Therefore, the value of an equivalent
1  1 1 1  single capacitor is 5.45 μF.
C    C 1  C 2  C 3   (3)

Capacitors in parallel
Therefore, Equation (3) is used to calculate In a parallel arrangement, all capacitors
the total capacitance of the three capacitors have the same potential difference across
connected in series. them, as shown in Figure 1.37. However,

Thus, if two capacitors are in series, then the charges for all capacitors are different.

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Physics Form 2 Final.indd 27 25/10/2025 10:25