Problem 40 - University Physics with Modern Physics

Question

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?

Step-by-Step Solution

Answer

The plates must be 1,500 kilometers apart.

Step 1: Understand the Electric Field Formula

The electric field \(E\) between two parallel plates is given by the formula \(E = \frac{V}{d}\), where \(V\) is the voltage across the plates and \(d\) is the distance between the plates.

Step 2: Rearrange the Formula to Solve for Distance

We want to find the distance \(d\), so we rearrange the formula to solve for \(d\). To do this, multiply both sides of the equation by \(d\) and then divide both sides by \(E\): \(d = \frac{V}{E}\).

Step 3: Substitute the Given Values

Substitute the given values into the formula. Here, \(V = 1.5 \, \text{V}\) and \(E = 1.0 \, \mu \text{V/m} = 1.0 \times 10^{-6} \, \text{V/m}\). Thus, \(d = \frac{1.5}{1.0 \times 10^{-6}}\).

Step 4: Calculate the Distance

Now calculate \(d\) using the substituted values: \(d = \frac{1.5}{1.0 \times 10^{-6}} = 1.5 \times 10^{6} = 1,500,000 \, \text{meters}\).

Step 5: Interpret the Result

The result, \(1,500,000 \, \text{meters}\), means that in order to produce an electric field of 1.0 \(\mu \text{V/m}\) with a 1.5V battery, the plates would have to be 1,500,000 meters, or 1,500 kilometers, apart.

Key Concepts

Voltage

Voltage is an essential concept in understanding electric fields. It represents the electric potential difference between two points. Imagine it like the pressure that pushes electric charges through a circuit. In this context, voltage is the force that makes an electric field possible between two plates.

• Measured in volts (V), it tells us how strong an electric field can be for a given distance.
• The greater the voltage, the stronger the electric field when other factors are constant.

Understanding voltage involves realizing how this potential difference can drive the motion of electric charges. In the problem, the voltage of the AA battery is used to determine how far apart plates must be to achieve a weak electric field. With a voltage of 1.5V, we see how even a small battery can influence electric fields over large distances.

Distance Between Plates

The distance between plates in a parallel plate setup is crucial for calculating the electric field formed between them. This distance, denoted by \(d\), is what determines how diluted or concentrated an electric field is given a fixed voltage.

• When plates are closer together, the electric field is stronger and more concentrated.
• If the plates are farther apart, the same voltage results in a weaker electric field.

In this problem, by rearranging the formula \(E = \frac{V}{d}\), we can solve for the distance \(d\). Substituting the known values provides insight into how spacing affects field strength. For the specified conditions, calculating \(d\) gives us a colossal distance of 1,500 kilometers, illustrating the extent needed to sustain such a weak electric field with a 1.5V battery.

Electric Field Sensitivity

Electric field sensitivity is a measure of how weak an electric field can be yet still detectable by an organism or an instrument. The concept highlights the remarkable sensitivity some creatures, like certain sharks, possess. These sharks can detect electric fields as weak as 1.0\(\mu \mathrm{V} / \mathrm{m}\), showcasing their highly advanced biological electroreception.

• 1.0\(\mu \mathrm{V} / \mathrm{m}\) is a microvolt per meter, indicating a very subtle and weak electric field.
• This sensitivity surpasses many artificial detection devices we're familiar with.

Understanding this level of sensitivity illustrates why calculations like those in the problem are relevant. They allow us to compare human engineering to natural systems and appreciate the precision and sensitivity present in nature.

Physics Problem Solving

Physics problem-solving often involves translating real-world scenarios into mathematical forms. The solution to this problem required understanding and applying the formula for an electric field between two plates.

• Identifying the given values: voltage and desired field strength.
• Rearranging the formula to solve for the unknown: distance between plates.
• Substituting values and performing calculations to find a sensible result.

What makes this exercise intriguing is that it merges conceptual knowledge with mathematical application. By solving it, students not only practice rearranging equations but also gain insights into how electric fields operate in diverse contexts. This reinforces the connection between physics formulas and real-world applications, deepening comprehension through practical examples.