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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.

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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.

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Most popular questions from this chapter

A solid conducting sphere has net positive charge and radius \(R =\) 0.400 m. At a point 1.20 m from the center of the sphere, the electric potential due to the charge on the sphere is 24.0 V. Assume that \(V = 0\) at an infinite distance from the sphere. What is the electric potential at the center of the sphere?

A helium ion (He\(^{++}\)) that comes within about 10 fm of the center of the nucleus of an atom in the sample may induce a nuclear reaction instead of simply scattering. Imagine a helium ion with a kinetic energy of 3.0 MeV heading straight toward an atom at rest in the sample. Assume that the atom stays fixed. What minimum charge can the nucleus of the atom have such that the helium ion gets no closer than 10 fm from the center of the atomic nucleus? (1 fm = 1 \(\times\) 10\(^{-15}\) m, and \(e\) is the magnitude of the charge of an electron or a proton.) (a) 2\(e\); (b) 11\(e\); (c) 20\(e\); (d) 22\(e\).

For a particular experiment, helium ions are to be given a kinetic energy of 3.0 MeV. What should the voltage at the center of the accelerator be, assuming that the ions start essentially at rest? (a) -3.0 MV; (b) +3.0 MV; (c) +1.5 MV; (d) +1.0 MV.

A ring of diameter 8.00 cm is fixed in place and carries a charge of \(+\)5.00 \(\mu\)C uniformly spread over its circumference. (a) How much work does it take to move a tiny \(+\)3.00-\(\mu\)C charged ball of mass 1.50 g from very far away to the center of the ring? (b) Is it necessary to take a path along the axis of the ring? Why? (c) If the ball is slightly displaced from the center of the ring, what will it do and what is the maximum speed it will reach?

A gold nucleus has a radius of 7.3 \(\times 10^{-15}\) m and a charge of \(+79e\). Through what voltage must an alpha particle, with charge \(+2e\), be accelerated so that it has just enough energy to reach a distance of 2.0 \(\times 10^{-14}\) m from the surface of a gold nucleus? (Assume that the gold nucleus remains stationary and can be treated as a point charge.)

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