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

Short Answer

Expert verified
The plates must be 1.5 million meters, or 1,500 km, apart to produce the weak electric field detectable by certain sharks.

Step by step solution

01

Given Parameters

The voltage \(V\) is given as 1.5 volts and the electric field \(E\) is given as \(1.0 \mu \mathrm{V} / \mathrm{m}\), or \(1.0 x 10^{-6} \mathrm{V} / \mathrm{m}\). The goal is to find the separation distance \(d\) between the plates.
02

Relate Electric Field and Voltage

From the formula for electric fields, transpose to isolate \(d\), so \(d = \frac{V}{E}\).
03

Substitute Values Into the Equation

Substitute the given values into the equation \(d = \frac{V}{E}\) to get \(d = \frac{1.5 V}{1.0 x 10^{-6} \mathrm{V} / \mathrm{m}}\).
04

Calculate the Distance

Perform the calculation to find the distance. The units of volts cancel out, leaving the distance in meters.

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Key Concepts

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

Electric Field Sensitivity
Sharks have an astonishing ability to detect electric fields, even as weak as 1.0 microvolts per meter! This extraordinary sensitivity allows them to perceive the faint electrical signals produced by prey or objects in their environment. To get a better idea of how subtle this field is, compare it to everyday situations. The electric field generated by common household activities or devices typically measures in volts per meter, which is significantly stronger. Understanding electric field sensitivity not only enhances our knowledge of sharks but also inspires developments in technology, such as sensitive detection equipment used in medical and scientific applications. By grasping the concept of electric field sensitivity, you can appreciate how certain creatures like sharks have adapted to utilize such delicate signals for survival.
Electric Potential Difference
The electric potential difference, more commonly known as voltage, is a fundamental concept in understanding electricity. It refers to the difference in electric potential energy between two points in a field. This potential difference is what drives electric charges to move, creating a flow of current.
In the exercise, a 1.5 V AA battery has an electric potential difference of 1.5 volts. This means the battery can push electric charges across a field with this strength. The unit of volts is crucial as it signifies the potential energy per unit charge available in the electric field.
It's fascinating how even a relatively low voltage, like 1.5 volts, can be used in scientific scenarios, such as creating tiny electric fields that mimic what certain animals can perceive. With these insights, you gain a practical understanding of how voltage impacts our daily lives, from powering simple devices to simulating natural phenomena.
Parallel Plate Capacitor
A parallel plate capacitor is a simple device but crucial in understanding electric fields. It consists of two conductive plates separated by a distance. When these plates are connected to a voltage source, such as the AA battery mentioned earlier, they create a uniform electric field in the space between them.
The relationship between voltage and electric field in capacitors is given by the formula: \(d = \frac{V}{E}\), where \(d\) is the distance between the plates, \(V\) is the potential difference, and \(E\) is the electric field. This device model is crucial because it helps to visualize and calculate how potential difference and separation affect the electric field.
In applications, parallel plate capacitors are fundamental because they demonstrate how fields can be manipulated and measured, making them vital in both educational and technological settings. By experimenting with these setups, you understand how to control and utilize electric fields, much like how sharks use their natural sensitivity to navigate the waters.

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Most popular questions from this chapter

A thin spherical shell with radius \(R_{1}=3.00 \mathrm{~cm}\) is concentric with a larger thin spherical shell with radius \(R_{2}=5.00 \mathrm{~cm}\). Both shells are made of insulating material. The smaller shell has charge \(q_{1}=+6.00 \mathrm{nC}\) distributed uniformly over its surface, and the larger shell has charge \(q_{2}=-9.00 \mathrm{nC}\) distributed uniformly over its surface. Take the electric potential to be zero at an infinite distance from both shells. (a) What is the electric potential due to the two shells at the fol- (ii) \(r=4.00 \mathrm{~cm}\) lowing distance from their common center: (i) \(r=0\) (iii) \(r=6.00 \mathrm{~cm} ?\) (b) What is the magnitude of the potential difference between the surfaces of the two shells? Which shell is at higher potential: the inner shell or the outer shell?

(a) How much excess charge must be placed on a copper sphere \(25.0 \mathrm{~cm}\) in diameter so that the potential of its center is \(3.75 \mathrm{kV} ?\) Take the point where \(V=0\) to be infinitely far from the sphere, (b) What is the potential of the sphere's surface?

An alpha particle with kinetic energy \(9.50 \mathrm{MeV}\) (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 point charge \(q_{1}=+5.00 \mu \mathrm{C}\) is held fixed in space. From a horizontal distance of \(6.00 \mathrm{~cm},\) a small sphere with mass \(4.00 \times 10^{-3} \mathrm{~kg}\) and charge \(q_{2}=+2.00 \mu \mathrm{C}\) is fired toward the fixed charge with an initial speed of \(40.0 \mathrm{~m} / \mathrm{s}\). Gravity can be neglected. What is the acceleration of the sphere at the instant when its speed is \(25.0 \mathrm{~m} / \mathrm{s} ?\)

A point charge \(q_{1}=+2.40 \mu \mathrm{C}\) is held stationary at the origin. A second point charge \(q_{2}=-4.30 \mu \mathrm{C}\) moves from the point \(x=0.150 \mathrm{~m}, y=0\) to the point \(x=0.250 \mathrm{~m}, y=0.250 \mathrm{~m} .\) How much work is done by the electric force on \(q_{2} ?\)

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