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(I) What is the magnitude of the electric field between two parallel plates \(4.0 \mathrm{~mm}\) apart if the potential difference between them is \(110 \mathrm{~V} ?\)

Short Answer

Expert verified
The magnitude of the electric field is 27500 V/m.

Step by step solution

01

Understand the Relationship

The relationship between electric field (E), potential difference (V), and distance (d) between plates is given by the formula: \[ E = \frac{V}{d} \] where E is the electric field in volts per meter (V/m), V is the potential difference in volts (V), and d is the distance in meters (m).
02

Convert Units

Firstly, convert the distance from millimeters to meters,\[ 4.0 \text{ mm} = 0.004 \text{ m}. \]
03

Apply the Formula

Use the formula for the electric field:\[ E = \frac{110 \text{ V}}{0.004 \text{ m}}. \]
04

Calculate the Electric Field

Carry out the calculation:\[ E = 27500 \text{ V/m}. \]

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Key Concepts

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

Understanding Potential Difference
Potential difference is a key concept in physics, particularly in the study of electric fields. It refers to the difference in electric potential between two points in an electric field.
The unit of potential difference is the volt (V). In simpler terms, potential difference measures how much work is needed to move a charge between two points.
It can be thought of as the "push" that drives electric charges through a circuit.
In the context of parallel plates, the potential difference is established across the two plates due to a voltage source. This means that one plate is at a higher potential than the other.
When this happens, an electric field is created in the region between the plates.
  • The larger the potential difference, the stronger the electric field that is produced.
  • A higher potential difference means more energy is available to move charges.
  • The potential difference is often created by connecting the plates to a battery or any voltage source.
Parallel Plates in Electric Fields
Parallel plates are often used in physics experiments and applications for generating uniform electric fields.
These are two flat surfaces placed parallel to each other, separated by a certain distance.
When a potential difference is applied to these plates, an electric field is established between them.
The field is considered uniform, meaning that its strength and direction are the same at every point between the plates.
  • The distance between the plates is crucial, as the electric field strength depends on it.
  • A smaller separation between the plates results in a stronger electric field for the same potential difference.
  • This setup is used in capacitors, where the plates store electric charge.
  • Common applications include devices like sensors, cameras, and scientific instruments.
Parallel plates offer a simple way to study the properties of electric fields and are used widely in educational settings to help students visualize electric fields in a controlled environment.
Electric Field Calculation Explained
The electric field is a vector quantity that represents the force experienced by a unit positive charge in the presence of an electric charge distribution.
When calculating the electric field between parallel plates, the formula used is \[E = \frac{V}{d}\]Here, \(E\) is the electric field, \(V\) is the potential difference, and \(d\) is the distance between the plates.
For accurate calculation, it is critical to have all measurements in the correct units—usually volts for potential difference and meters for distance.
  • First, convert any given measurements to their standard units. For example, convert millimeters to meters by dividing by 1000.
  • Insert these values into the formula to solve for \(E\).
  • Finally, carry out the division to find the electric field in volts per meter (V/m).
The formula shows that increasing the potential difference while keeping the distance constant results in a stronger electric field.
Understanding this relationship is crucial for accurately describing how electric fields behave in various situations.
It also illustrates the practical importance of precision in measuring and calculating electrical quantities, as even small changes in measurement can significantly impact the resulting electric field.

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Most popular questions from this chapter

(II) An electron starting from rest acquires \(1.33 \mathrm{keV}\) of kinetic energy in moving from point A to point B. (a) How much kinetic energy would a proton acquire, starting from rest at \(B\) and moving to point \(A ?(b)\) Determine the ratio of their speeds at the end of their respective trajectories.

Sketch the electric field and equipotential lines for two charges of the same sign and magnitude separated by a distance \(d\).

(II) The Earth produces an inwardly directed electric field of magnitude \(150 \mathrm{~V} / \mathrm{m}\) near its surface. \((a)\) What is the potential of the Earth's surface relative to \(V=0\) at \(r=\infty ?(b)\) If the potential of the Earth is chosen to be zero, what is the potential at infinity? (Ignore the fact that positive charge in the ionosphere approximately cancels the Earth's net charge; how would this affect your answer?)

(II) The electric potential of a very large isolated flat metal plate is \(V_{0} .\) It carries a uniform distribution of charge of surface density \(\sigma\left(\mathrm{C} / \mathrm{m}^{2}\right),\) or \(\sigma / 2\) on each surface. Determine \(V\) at a distance \(x\) from the plate. Consider the point \(x\) to be far from the edges and assume \(x\) is much smaller than the plate dimensions.

(II) A dust particle with mass of \(0.050 \mathrm{~g}\) and a charge of \(2.0 \times 10^{-6} \mathrm{C}\) is in a region of space where the potential is given by \(V(x)=\left(2.0 \mathrm{~V} / \mathrm{m}^{2}\right) x^{2}-\left(3.0 \mathrm{~V} / \mathrm{m}^{3}\right) x^{3} .\) If the particle starts at \(x=2.0 \mathrm{~m},\) what is the initial acceleration of the charge?

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