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Two identical metal plates are given positive charge \(Q_{1}\) and \(Q_{2}\left(Q_{2}

Short Answer

Expert verified
(c) \(\frac{Q_{1}-Q_{2}}{C}\)

Step by step solution

01

Understanding the Charge Redistribution

When the two metal plates are brought together to form a capacitor, charge redistribution occurs so that the potential difference across the plates is uniform. If we assume the final charges on the plates are \(+q\) and \(-q\), then the net charge of the system must be conserved, leading to the equation: \[ Q_1 + Q_2 = q + (-q) \] This simplifies to \(+q - q = 0\), allowing focus on potential difference.
02

Calculate the Potential Difference

The potential difference \(V\) across the plates of a capacitor is determined by the formula: \[ V = \frac{Q}{C} \] where \(Q\) is the charge on one plate and \(C\) is the capacitance. Since charge is redistributed, the effective charge difference that contributes to the potential difference is \(Q_1 - Q_2\). Thus, the potential difference becomes: \[ V = \frac{Q_1 - Q_2}{C} \] This is the formula for the potential difference between two parallel plates.

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Key Concepts

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

Charge Redistribution
In any capacitor system, charge redistribution is a crucial process. This occurs when two or more charged objects, such as metal plates, come into close proximity and need to form a stable system. In our scenario, we begin with two metal plates holding different charges, denoted as \(Q_1\) and \(Q_2\), where \(Q_2 < Q_1\). When these plates are arranged as a parallel plate capacitor, the charges will redistribute in a way that makes the electric potential difference across them uniform.

This means that the charges flow so that one plate becomes positively charged at \(+q\) and the other negatively charged at \(-q\), achieving a net balance. Through this process, the system conserves the total initial charge. Even though \(Q_1 + Q_2\) initially describes the system, after redistribution, the combination of \(+q\) and \(-q\) ensures that the total effective charge results in a net potential difference.

Charge redistribution helps ensure a stable configuration for the capacitor, allowing it to function effectively as an energy storage device.
Parallel Plate Capacitor
A simple yet effective design, the parallel plate capacitor is one of the most common types of capacitors used in electronic circuits. It consists of two conductive plates placed parallel and separated by a distance, often filled with an insulating material called a dielectric. The arrangement allows the capacitor to store electric energy by maintaining a surface charge on each plate.

When these capacitor plates are charged, each holds a large number of equal and opposite charges, creating an electric field between them. The strength of the electric field depends on the amount of charge and the distance between the plates. The uniform electric field between the plates is a key feature, simplifying calculations for potential difference and capacitance.

This setup is particularly favored for its simplicity in design and manufacturing, making it an essential component in many circuits that require energy storage, filtering, or timing functions.
  • The design ensures a uniform electric field.
  • It allows for easy calculation of potential differences.
  • The capacitor's effectiveness is influenced by the plate area and separation distance.
Capacitor Equations
Understanding the equations governing capacitors is fundamental to grasping how they work. The main concepts include the relation between charge \(Q\), capacitance \(C\), and potential difference \(V\). For any capacitor:

- The capacitance \(C\) is defined by the formula \(C = \frac{Q}{V}\), symbolizing the capacitor's ability to store charge per unit potential difference.- The potential difference \(V\) across the capacitor is described by \(V = \frac{Q}{C}\). This is key when assessing the voltage required for a given charge storage.

For parallel plate capacitors, as in our example, the charges \(Q_1\) and \(Q_2\) initially affect the system. Upon forming a capacitor, the equation for potential difference adapts to account for charge redistribution as \(V = \frac{Q_1 - Q_2}{C}\). This formula applies particularly when only the difference in initial charges contributes to the voltage potential.
  • Capacitance is a measure of charge storage efficiency.
  • Potential difference determines the energy transfer potential.
  • Equations simplify calculation of electrical characteristics in circuits.

Understanding these fundamental relationships empowers students to accurately predict how capacitors will behave under various conditions, making them crucial tools in electronics design.

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Most popular questions from this chapter

Three identical capacitors are first connected in series and then first and last conductors of combination are connected to the earth. A charge \(Q\) is given to second conductor of first capacitor. Then potential of this conductor is : (a) \(\frac{Q}{3 C}\) (b) \(\frac{Q}{C}\) (c) \(\frac{2 Q}{3 C}\) (d) \(\frac{3 Q}{2 C}\)

The capacitance (C) for an isolated conducting sphere of radius \((a)\) is given by \(4 \pi \varepsilon_{0} a .\) If the sphere is enclosed with an earthed concentric sphere, the ratio of the radii of the spheres being \(\frac{n}{(n-1)}\) then the capacitance of such a sphere will be increased by a factor? (a) \(n\) (b) \(\frac{n}{(n-1)}\) (c) \(\frac{(n-1)}{n}\) (d) \(a n\)

Inside two identical capacitors, two identical dielectric slabs are introduced as shown in figure. What will happen, if slab of capacitor \(B\) is pulled out, with the battery remain connected ? (a) During the process charge flows from \(a\) to \(b\) (b) Finally charge on \(B\) will be less than charge on \(A\) (c) During the process work done by external force \(F\) appear as heat in the circuit (d) None of the above

An isolated metallic object is charged in vacuum to a potential \(V_{0}\), its electrostatic energy being \(W_{0}\). It is then disconnected from the source of potential, its charge being left unchanged and is immersed in a large volume of dielectric, with dielectric constant \(K\). The electrostatic energy will be (a) \(K W_{0}\) (b) \(W_{0} / K\) (c) \(\frac{W_{0}}{2 K}\) (d) \(W_{0}\)

The amount of charge flow, when a conducting sphere of radius \(R\) and carrying a charge \(Q\), is joined to an uncharged conducting sphere of radius \(2 R\) is : (a) \(\frac{Q}{4}\) (b) \(\frac{Q}{3}\) (c) \(\frac{Q}{2}\) (d) \(\frac{2 Q}{3}\)

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