Problem

If $C_1 = 14 \, \mathrm{\mu F}$, $C_2 = 20 \, \mathrm{\mu F}$, and $C_3 = 12 \, \mathrm{\mu F}$, with the insulated plate of $C_1$ at $100 \, \mathrm{V}$, and one plate of $C_3$ grounded, what is the potential difference between the plates of $C_2$, given that all capacitors are in series?

Data

• Capacitor $C_1 = 14 \, \mathrm{\mu F} = 14 \times 10^{-6} \, \mathrm{F}$
• Capacitor $C_2 = 20 \, \mathrm{\mu F} = 20 \times 10^{-6} \, \mathrm{F}$
• Capacitor $C_3 = 12 \, \mathrm{\mu F} = 12 \times 10^{-6} \, \mathrm{F}$
• Voltage $V = 100 \, \mathrm{V}$.

Prerequisite Concepts

• Series Capacitance:

$$\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$$

• Charge Conservation in Series:
  $Q_1 = Q_2 = Q_3 = Q$.
• Voltage Across Each Capacitor:

$$V_i = \frac{Q}{C_i}$$

Answer

Step 1: Calculate Total Capacitance

$$\frac{1}{C_{\text{eq}}} = \frac{1}{14} + \frac{1}{20} + \frac{1}{12}$$

Take LCM:

$$\frac{1}{C_{\text{eq}}} = \frac{30 + 21 + 35}{420} = \frac{86}{420}$$ $$C_{\text{eq}} = \frac{420}{86} = 4.883 \, \mathrm{\mu F}$$

Step 2: Calculate Total Charge

$$Q = C_{\text{eq}} \cdot V$$ $$Q = 4.883 \times 10^{-6} \cdot 100 = 488.3 \, \mathrm{\mu C}$$

Step 3: Calculate Voltage Across $C_2$

$$V_2 = \frac{Q}{C_2}$$ $$V_2 = \frac{488.3}{20} = 24.42 \, \mathrm{V}$$

Final Answer

The potential difference across $C_2$ is $24.42 \, \mathrm{V}$.

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