/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 21 Two conducting plates \(x\) and ... [FREE SOLUTION] | 91Ó°ÊÓ

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Two conducting plates \(x\) and \(y\), each having large surface area \(A\) (on the side) are placed parallel to each other. The plate \(x\) is given a charge \(Q\) whereas the other is neutral. The electric field at a point in between the plates is given by: (a) \(\frac{Q}{2 A}\) (b) \(\frac{Q}{2 A \varepsilon_{0}}\) towards left (c) \(\frac{Q}{2 A \varepsilon_{0}}\) towards right (d) \(\frac{Q}{2 \varepsilon_{0}}\) towards right

Short Answer

Expert verified
Option (c): \(\frac{Q}{2 A \varepsilon_{0}}\) towards right.

Step by step solution

01

Understanding the Problem

We have two conducting plates, plate \( x \) with charge \( Q \) and plate \( y \), which is neutral. We are to find the electric field between the plates and identify its direction based on given options.
02

Determine Electric Field Due to a Single Charged Plate

The electric field due to a large charged conductor plate with charge density \( \sigma \) is given by \( E = \frac{\sigma}{2 \varepsilon_{0}} \). Here, the charge density \( \sigma = \frac{Q}{A} \), so the electric field \( E = \frac{Q}{2 A \varepsilon_{0}} \).
03

Determine the Overall Electric Field

Plate \( y \), being neutral, has a zero net electric field impact. Hence, only the field due to plate \( x \) is considered. Thus, the electric field between the plates is \( \frac{Q}{2 A \varepsilon_{0}} \). Because electric field lines originate from positive charges, the field direction is away from plate \( x \) which has charge \( Q \), towards the right.
04

Select the Correct Answer

Looking at our calculation, the correct answer is option (c), which states the electric field is \( \frac{Q}{2 A \varepsilon_{0}} \) towards right.

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Key Concepts

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

Charge Density
Charge density is a measure of the amount of electric charge per unit area on a conducting surface. When looking at parallel conducting plates, charge density becomes a crucial factor in understanding how electric fields develop between the plates. For a plate with a total charge \( Q \) and surface area \( A \), the surface charge density \( \sigma \) can be calculated by:
  • \( \sigma = \frac{Q}{A} \)
This equation tells us that the charge density is directly proportional to the total charge applied to the plate and inversely proportional to the surface area of the plate. So, if you increase the charge, the charge density increases. Conversely, if you increase the area while keeping the charge constant, the charge density decreases. By understanding charge density, we can easily determine how strong an electric field will be emanating from a charged surface. This concept is foundational when examining the behavior of electric fields between conducting plates.
Gauss's Law
Gauss's Law connects the electric field and the distribution of charge across a surface. Simply put, it relates the electric flux through a closed surface to the charge enclosed by that surface. The law is expressed mathematically as:
  • \( \Phi = \oint E \cdot dA = \frac{Q_{encl}}{\varepsilon_{0}} \)
In the context of large conducting plates, Gauss's Law is instrumental in calculating the electric field. By employing Gaussian surfaces, typically a cylindrical surface through the charged plate, it becomes feasible to find the electric field vector \( E \). Since the electric field direction is always perpendicular to the surface of the conductor, we use Gauss's Law to determine that the electric field strength is \( E = \frac{\sigma}{2 \varepsilon_{0}} \) due to a single plate. Knowing where the field originates clarifies the behavior of electric fields between conducting plates.
Conducting Plates
Conducting plates are metal plates capable of conducting electricity and interacting with electric fields. These plates, when charged, influence the electric field in their vicinity. For two parallel conducting plates, one charged and the other neutral, the arrangement establishes a uniform electric field between them. The charged plate creates an electric field that impacts only the region between the plates, given that fields don't affect areas outside of infinite parallel plates significantly. This is because the charges disperse symmetrically owing to the plate's significant surface area, creating a constant electric field. It is crucial to remember that conducting plates are assumed to have negligible resistance such that they don't oppose the electric field formed.
Electrostatics
Electrostatics studies electric fields, forces, and potentials in systems with stationary charges. In this realm, concepts such as charge density, electric field, and potential energy play prominent roles. When analyzing systems like conducting plates, electrostatics allows us to predict how charges will organize themselves and how resulting electric fields behave. Specifically, in a system with conducting plates, we learn that:
  • The fields between plates are the primary concern for calculating electric forces.
  • Electric field lines run from positive to negative charges, dictating field directions.
  • Neutral plates do not impact field magnitude, as their net charge is zero.
Through electrostatics, we understand that fields originate from charged surfaces and, in arrangements like this, are uniform across the space between the plates. Such clarity in behavior makes evaluating potential effects on objects placed between conducting plates manageable and predictable.

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Most popular questions from this chapter

Two capacitors of equal capacity are connected in series, they have some resultant capacity. Now they are connected in parallel. The resultant capacity becomes: (a) four times of the previous value (b) one-fourth of the previous value (c) twice of the previous value (d) half of the previous value

A capacitor of capacitance \(C\) is charged to a potential difference \(V\) from a cell and then disconnected from it. A charge \(+Q\) is now given to its positive plate. The potential difference across the capacitor is now: (a) \(V\) (b) \(V+\frac{Q}{C}\) (c) \(V+\frac{\partial}{2 C}\) (d) \(V-\frac{Q}{C}\)

A capacitor of capacitance \(C\) is charged to a potential difference \(V_{0}\). The charged battery is disconnected and the capacitor is connected to a capacitor of unknown capacitance \(C_{x}\). The potential difference across the combination is \(V\). The value of \(C_{r}\) should be : (a) \(\frac{C\left(V_{0}-V\right)}{V}\) (b) \(\frac{C\left(V-V_{0}\right)}{V}\) (c) \(\frac{\mathrm{CV}}{V_{0}}\) (d) \(\frac{C V_{0}}{V}\)

A parallel plate capacitor of capacitance \(C\) is connected across a battery of emf. If the separation between the plates is doubled, the force of attraction between the plates \(\ldots \ldots \ldots\) by a factor of \(\ldots \ldots \ldots\) : (a) increases, 4 (b) decreases, 4 (c) decreases, 2 (d) remains same

A capacitor is charged by using a battery, which is then disconnected. A dielectric slab is then slided between the plates which results in: (a) reduction of charge on the plates and increase of potential difference across the plates (b) increase in the potential difference across the plates, reduction in stored energy, but no change in the charge on the plates (c) decrease in the potential difference across the plates, reduction in stored energy, but no change in the charge on the plates (d) none of the above

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