/*! 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 38 Certain sharks can detect an ele... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Certain sharks can detect an electric field as weak as 1.0 \(\mu\)V\(/\)m. To grasp how weak this field is, if you wanted to produce it between two parallel metal plates by connecting an ordinary 1.5V AA battery across these plates, how far apart would the plates have to be?

Short Answer

Expert verified
The plates need to be 1.5 million meters apart.

Step by step solution

01

Understanding the Electric Field Equation

The electric field (E) created between two parallel plates connected to a battery can be described using the equation \( E = \frac{V}{d} \), where \( V \) is the voltage of the battery, and \( d \) is the distance between the plates. In our case, \( V = 1.5 \, V \) and \( E = 1.0 \, \mu V/m = 1.0 \times 10^{-6} \, V/m \).
02

Rearranging the Formula for Distance

To find the distance \( d \), we need to rearrange the formula \( E = \frac{V}{d} \). Solving for \( d \), we get \( d = \frac{V}{E} \).
03

Substitute Known Values

Substitute the known values into the rearranged formula: \(d = \frac{1.5 \, V}{1.0 \times 10^{-6} \, V/m} \).
04

Calculate the Distance

Perform the calculation: \( d = \frac{1.5}{1.0 \times 10^{-6}} = 1.5 \times 10^6 \, m \). The plates need to be \(1.5 \times 10^6\) meters apart.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Voltage
Voltage is the measure of electric potential difference between two points. In an electronic circuit, it indicates how much electric potential energy is available to push electrons through the circuit. This potential difference, measured in volts (V), is like the driving force that moves through the circuit.
In the context of parallel plates, the voltage of a battery, such as a 1.5V AA battery, is used. This is what creates an electric field between the plates when connected. Voltage tells us how strong the electrical "push" or "pull" between the plates is.
Understanding voltage helps us arrange components properly to achieve desired electric fields and effects within circuits.
Parallel Plates
Parallel plates are two conductive sheets positioned to face each other, usually in a parallel manner, separated by a certain distance. They often serve as a technique for creating stable electric fields.
When connected to a voltage source like a battery, the parallel plates manage to create a uniform electric field in the space between them. This field is constant in strength and direction across the entire area between the plates, which makes them very useful in experiments and various electronics like capacitors.
The field between parallel plates follows the formula: \[ E = \frac{V}{d} \] where \( E \) is the electric field, \( V \) is the voltage applied, and \( d \) signifies the separation distance.
Distance Calculation
Calculating the distance between parallel plates can require solving the electric field equation for distance. In our exercise, we want to find out how far apart the plates need to be to create a precise electric field.
To do this, you can rearrange the electric field formula: \[ E = \frac{V}{d} \] to solve for distance \( d \): \[ d = \frac{V}{E} \] Once rearranged, you can substitute in the known values of voltage and the desired electric field strength.
The calculation in the exercise uses a standard AA battery voltage of 1.5V and a target electric field of \(1.0 \times 10^{-6} \, V/m\), resulting in a very large distance (\(1.5 \times 10^6\) meters) to achieve such a weak field.
Electric Field Detection
Electric field detection involves sensing the presence and strength of an electric field in an area. Certain animals, like sharks, have the natural ability to detect these fields, even when they are extremely weak.
In physics and engineering, being able to detect an electric field allows for understanding how charges and voltages distribute around various devices.
The field is detected by its effect on charged particles or electronic sensors. The amount of force experienced by a charged particle in a field reveals the strength of that field. This detection is critical in measuring and understanding electric phenomena in diverse applications such as sensors and electrostatic experiments.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Point charges \(q_1 = +\)2.00 \(\mu\)C and \(q_2 = -\)2.00 \(\mu\)C are placed at adjacent corners of a square for which the length of each side is 3.00 cm. Point \(a\) is at the center of the square, and point \(b\) is at the empty corner closest to \(q_2\) . Take the electric potential to be zero at a distance far from both charges. (a) What is the electric potential at point a due to \(q_1\) and \(q_2\)? (b) What is the electric potential at point \(b\)? (c) A point charge \(q_3 = -\)5.00 \(\mu\)C moves from point \(a\) to point \(b\). How much work is done on \(q_3\) by the electric forces exerted by \(q_1\) and \(q_2\)? Is this work positive or negative?

Two large, parallel conducting plates carrying opposite charges of equal magnitude are separated by 2.20 cm. (a) If the surface charge density for each plate has magnitude 47.0 nC\(/m^2\), what is the magnitude of \(\overrightarrow{E}\) in the region between the plates? (b) What is the potential difference between the two plates? (c) If the separation between the plates is doubled while the surface charge density is kept constant at the value in part (a), what happens to the magnitude of the electric field and to the potential difference?

A positive point charge \(q_1 = +5.00 \times 10^{-4}\) C is held at a fixed position. A small object with mass 4.00 \(\times 10^{-3}\) kg and charge \(q_2 = -3.00 \times 10^{-4}\) C is projected directly at \(q_1\) . Ignore gravity. When \(q_2\) is 0.400 m away, its speed is 800 m\(/\)s. What is its speed when it is 0.200 m from \(q_1\) ?

An alpha particle with kinetic energy 9.50 MeV (when far away) collides head- on with a lead nucleus at rest. What is the distance of closest approach of the two particles? (Assume that the lead nucleus remains stationary and may be treated as a point charge. The atomic number of lead is 82. The alpha particle is a helium nucleus, with atomic number 2.)

A total electric charge of 3.50 nC is distributed uniformly over the surface of a metal sphere with a radius of 24.0 cm. If the potential is zero at a point at infinity, find the value of the potential at the following distances from the center of the sphere: (a) 48.0 cm; (b) 24.0 cm; (c) 12.0 cm.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.