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Chapter 17: Problem 60

A shark is able to detect the presence of electric fields as small as $1.0 \mu \mathrm{V} / \mathrm{m} .$ To get an idea of the magnitude of this field, suppose you have a parallel plate capacitor connected to a 1.5 - \(V\) battery. How far apart must the parallel plates be to have an electric field of $1.0 \mu \mathrm{V} / \mathrm{m}$ between the plates?

Short Answer

Expert verified
Answer: The distance between the parallel plates must be 1.5 x 10^6 meters.

Step by step solution

01

Write down the formula for electric field in a parallel plate capacitor

The electric field in a parallel plate capacitor can be calculated using the following formula: \(E = \frac{V}{d}\) where \(E\) is the electric field, \(V\) is the voltage, and \(d\) is the distance between the plates.
02

Plug in the given values

We are given the electric field \(E = 1.0 \mu \mathrm{V} / \mathrm{m}\) and the voltage \(V = 1.5 \, \mathrm{V}\). We can plug these values into the formula: \(1.0 \times 10^{-6} \, \mathrm{V/m} = \frac{1.5 \, \mathrm{V}}{d}\)
03

Solve for the distance \(d\)

Now we need to solve for \(d\): \(d = \frac{1.5 \, \mathrm{V}}{1.0 \times 10^{-6} \, \mathrm{V/m}}\)
04

Calculate the distance

Multiply both the numerator and denominator: \(d = \frac{1.5}{1.0 \times 10^{-6}} \, \mathrm{m}\) \(d = 1.5 \times 10^{6} \, \mathrm{m}\)
05

Write the final answer

The distance between the parallel plates must be \(1.5\times 10^{6}\) meters to have an electric field of \(1.0 \mu \mathrm{V/m}\).

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Most popular questions from this chapter

A positively charged oil drop is injected into a region of uniform electric field between two oppositely charged, horizontally oriented plates spaced $16 \mathrm{cm}$ apart. If the electric force on the drop is found to be $9.6 \times 10^{-16} \mathrm{N}\( and the potential difference between the plates is \)480 \mathrm{V}$ what is the magnitude of the charge on the drop in terms of the elementary charge \(e\) ? Ignore the small buoyant force on the drop.
In \(1911,\) Ernest Rutherford discovered the nucleus of the atom by observing the scattering of helium nuclei from gold nuclei. If a helium nucleus with a mass of \(6.68 \times 10^{-27} \mathrm{kg},\) a charge of \(+2 e,\) and an initial velocity of \(1.50 \times 10^{7} \mathrm{m} / \mathrm{s}\) is projected head-on toward a gold nucleus with a charge of \(+79 e\), how close will the helium atom come to the gold nucleus before it stops and turns around? (Assume the gold nucleus is held in place by other gold atoms and does not move.)
A parallel plate capacitor has a charge of \(0.020 \mu \mathrm{C}\) on each plate with a potential difference of \(240 \mathrm{V}\). The parallel plates are separated by \(0.40 \mathrm{mm}\) of air. (a) What is the capacitance for this capacitor? (b) What is the area of a single plate? (c) At what voltage will the air between the plates become ionized? Assume a dielectric strength of \(3.0 \mathrm{kV} / \mathrm{mm}\) for air.
A large parallel plate capacitor has plate separation of \(1.00 \mathrm{cm}\) and plate area of \(314 \mathrm{cm}^{2} .\) The capacitor is connected across a voltage of \(20.0 \mathrm{V}\) and has air between the plates. How much work is done on the capacitor as the plate separation is increased to $2.00 \mathrm{cm} ?$
Two metal spheres have charges of equal magnitude, $3.2 \times 10^{-14} \mathrm{C},$ but opposite sign. If the potential difference between the two spheres is \(4.0 \mathrm{mV},\) what is the capacitance? [Hint: The "plates" are not parallel, but the definition of capacitance holds.]
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