محاضرة 11 فيزياء عامة (2) كهربية ساكنة أمثلة محلولة عن المكثف الكهربي
أمثلة محلولة عن المكثف الكهربي
An air-filled capacitor consists of two plates, each with an area of 7.6cm2, separated by a distance of 1.8mm. If a 20V potential difference is applied to these plates, calculate,
- (a) the electric field between the plates,
- (b) the surface charge density,
- (c) the capacitance, and
- (d) the charge on each plate.
Find the equivalent capacitance between points a and b for the group of capacitors shown in figure 6.7 . C1=1mF, C2=2mF, C3=3mF, C4=4mF, C5=5mF, and C6=6mF.
First the capacitor C3 and C6 are connected in series so that the equivalent capacitance Cde is
Second C1 and C5 are connected in parallel
The circuit become as shown below
Continue with the same way to reduce the circuit for the capacitor C2 and Cde to get Cgh=4mF
Capacitors Cmg and Cgh are connected in series the result is Cmh=2mF, The circuit become as shown below
Capacitors Cmh and Ckl are connected in parallel the result is
In the above example 6.3 determine the potential difference across each capacitor and the charge on each capacitor if the total charge on all the six capacitors is 384mC.
First consider the equivalent capacitor Ceq to find the potential between points a and b (Vab)
Second notice that the potential Vkl=Vab since the two capacitors between k and lare in parallel, the potential across the capacitors C1 and C5 = 48V.
V1=48V and Q1=C1V1=48mC
And for C5
V5=48V and Q5=C5V5=240mC
For the circuit (iv) notice that Vmh=Vab=48V, and
Since the two capacitors shown in the circuit (iii) between points m and h are in series, each will have the same charge as that of the equivalent capacitor, i.e.
Therefore for C4, V4=24 and Q4=96mC
In the circuit (ii) the two capacitor between points g and h are in parallel so the potential difference across each is 24V.
Therefore for C2, V2=24V and Q2=C2V2=48mC
Also in circuit (ii) the potential difference
The two capacitors shown in circuit (i) between points d and a are in series, and therefore the charge on each is equal to Qde.
Therefore for C6, Q6=48mC
For C3, Q3=48mC and V3=Q3/C3=16V
The results can be summarized as follow: