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Two capacitors, \(3 \mu \mathrm{F}\) and \(4 \mu \mathrm{F}\), are individually charged across a 6 volt battery. After being disconnected from the battery, they are connected together with the negative plate of one attached to the positive plate of the other. What is the common potential? (a) 6 volt (b) \((6 / 7)\) volt (c) 2 volt (d) none of these

Short Answer

Expert verified
None of the given options; the common potential is 3.5 volts.

Step by step solution

01

Determine Initial Charge on Capacitors

First, calculate the charge initially stored on each capacitor when connected to a 6 volt battery. The formula to find the charge, \( Q \), is given by:\[ Q = C \times V \]For the 3 \( \mu \mathrm{F} \) capacitor, \( Q_1 = 3 \times 6 = 18 \mu \mathrm{C} \).For the 4 \( \mu \mathrm{F} \) capacitor, \( Q_2 = 4 \times 6 = 24 \mu \mathrm{C} \).
02

Analyze Charge Redistribution

When the capacitors are connected together, the total charge is conserved. The charges on the capacitors will redistribute as they are connected with opposite polarities. Thus, total charge, \( Q_t \), shared by both is:\[ Q_t = Q_2 - Q_1 = 24 \mu \mathrm{C} - 18 \mu \mathrm{C} = 6 \mu \mathrm{C} \].
03

Calculate Common Potential

The common potential (\( V_c \)) across both capacitors can be found using the total capacitance of the series combination. The total capacitance, \( C_t \), for capacitors in series is calculated as:\[ \frac{1}{C_t} = \frac{1}{3 \mu \mathrm{F}} + \frac{1}{4 \mu \mathrm{F}} = \frac{4}{12} + \frac{3}{12} = \frac{7}{12} \]Thus, \( C_t = \frac{12}{7} \mu \mathrm{F} \).Using the formula \( V_c = \frac{Q_t}{C_t} \):\[ V_c = \frac{6 \mu \mathrm{C}}{\frac{12}{7} \mu \mathrm{F}} = \frac{6 \times 7}{12} = \frac{42}{12} = 3.5 \text{ volts} \]
04

Evaluate Against Options

Compare the calculated common potential of 3.5 volts against the given options. None of the options (a) 6 volt, (b) \((6 / 7)\) volt, (c) 2 volt exactly match the value of 3.5 volts.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Capacitance
Capacitance is a fundamental concept in electrostatics that allows us to understand how capacitors store electric charge. A capacitor is a device capable of storing electrical energy by separating charges across two conductive plates. The ability of a capacitor to store charge for a given potential difference (voltage) is determined by its capacitance.The capacitance, denoted as \(C\), is defined as the ratio of the charge \(Q\) stored on one plate to the voltage \(V\) across the plates, represented mathematically as:\[ C = \frac{Q}{V} \]This formula implies that capacitance is directly proportional to the amount of charge a capacitor can hold for a particular voltage. Larger capacitance values indicate that more charge can be stored at a particular voltage.
  • Capacitance is measured in farads (F), often expressed in microfarads (\(\mu \mathrm{F}\)), nanofarads (nF), or picofarads (pF) because the farad is a very large unit.
  • Factors affecting capacitance include plate area, distance between plates, and the dielectric material between the plates.
  • A real-world example of capacitance can be found in electronic gadgets like phone chargers and stereo systems, where capacitors play a crucial role in maintaining electronic stability.
Charge redistribution
When capacitors are connected after being charged, the charges on their plates can change unless they are isolated. This phenomenon is known as charge redistribution. It occurs when two or more capacitors are connected in such a way that they share and equilibrate their stored charges. In our example, two different capacitors initially charged across a battery are disconnected and then interconnected with opposite polarities. This setup causes charge to flow between them until they reach a common potential.
  • The total initial charge available remains constant, according to the law of conservation of charge.
  • The charge rearranges itself between the capacitors such that the final potential across both is the same.
  • This redistribution is crucial for achieving a stable electrical state between the capacitors.
By ensuring that the final voltage across each capacitor becomes the same, charge redistribution allows us to apply electrostatic principles to solve complex capacitor networks.
Potential difference
The potential difference, commonly referred to as voltage, is the difference in electric potential between two points in an electric field. It is the work needed per unit charge to move a test charge between these two points. Voltage, given in volts (V), plays a critical role in the behavior of capacitors, as it drives the movement and storage of charge. In capacitors, voltage not only affects the amount of charge stored but also determines how energy is stored and released.
  • Potential difference is proportional to the electric field strength between two points. The stronger the electric field, the higher the potential difference.
  • A practical example of potential difference is found in battery-driven devices. The battery creates a potential difference, which causes current to flow through the circuit.
  • The concept of potential difference is crucial for understanding why charges redistribute to achieve equilibrium in connected capacitors.
In scenarios like the one described in the exercise, connecting capacitors causes them to seek a common potential, transferring charge until all interconnected capacitors experience the same voltage. This redistribution leads to new equilibrium states, showcasing the importance of potential difference in electrostatic systems.

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