فيزياء عامة (2)

# محاضرة 11 فيزياء عامة (2) كهربية ساكنة أمثلة محلولة عن المكثف الكهربي

## أمثلة محلولة عن المكثف الكهربي

Example 6.1

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.

Solution

اعلانات جوجل Example 6.3

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. Solution

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

Ckl=1+5=6mF

اعلانات جوجل

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

اعلانات جوجل Ceq=8mF

Example 6.4

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.

اعلانات جوجل

Solution

First consider the equivalent capacitor Ceq to find the potential between points a and b (Vab)

اعلانات جوجل 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

Qmh=CmhVmh=2×28=96mC 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.

Qmh=Qgh=Qmh=96mC

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

Vde=Vgh=24V

###### And

Qde=CdeVde=2×24=48mC

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: ### الدكتور حازم فلاح سكيك

د. حازم فلاح سكيك استاذ الفيزياء المشارك في قسم الفيزياء في جامعة الازهر – غزة | مؤسس شبكة الفيزياء التعليمية | واكاديمية الفيزياء للتعليم الالكتروني | ومنتدى الفيزياء التعليمي

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