Ratio of Capacitors

PhysicsCapacitanceEasySTEM

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The capacitance of a parallel plate capacitor is given by $C = \frac{\varepsilon A}{d}$. Given two capacitors:

1) Left capacitor: Dielectric constant $2\varepsilon$, area $4A$, distance $d$, labeled $C_1$.

2) Right capacitor: Dielectric constant $\varepsilon$, area $3A$, distance $2d$, labeled $C_2$.

Find the ratio: $\frac{C_1}{C_2} = ?$

This question includes visual content: The image shows two parallel plate capacitor diagrams drawn on grid paper. The left capacitor (labeled C1 below) has dielectric constant 2ε, plate area 4A, and separation distance d. The right capacitor (labeled C2 below) has dielectric constant ε, plate area 3A, and separation distance 2d. The question asks to find the ratio C1/C2.

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Step by Step Written Solution

1
Step 1

Merhaba süleyman, bu fizik sorusunu birlikte inceleyelim. İki farklı paralel plakalı kondansatörün sığa oranını bulmamız isteniyor.

Sığa Hesabı

2
Step 2

Öncelikle genel sığa formülünü hatırlayalım. Sığa, epsilon çarpı alan bölü uzaklık formülü ile hesaplanır.

$$C = \frac{\varepsilon A}{d}$$
3
Step 3

İlk kondansatör olan C bir için değerleri yazalım. Dielektrik katsayısı iki epsilon, yüzey alanı dört A ve plakalar arası uzaklık d kadardır.

$$C_1 = \frac{(2\varepsilon)(4A)}{d} = 8 \frac{\varepsilon A}{d}$$
4
Step 4

Şimdi ikinci kondansatör olan C iki için değerleri belirleyelim. Burada dielektrik katsayısı epsilon, alan üç A ve uzaklık iki d kadardır.

$$C_2 = \frac{(\varepsilon)(3A)}{2d} = 1,5 \frac{\varepsilon A}{d}$$

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About This Question

Subject
Physics
Topic
Capacitance
Difficulty
Easy
Exam
STEM

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