Ch 23: Electric Potential

Young & Freedman Calc - University Physics 14th Edition

Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook

Young & Freedman Calc 14th Edition

Ch 23: Electric Potential

Problem 38

Chapter 23, Problem 38

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

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Understand the relationship between electric field (E), voltage (V), and distance (d) between plates. The electric field between two parallel plates is given by the formula:

E = \frac{V}{d}

Identify the given values in the problem: the electric field (E) is 1.0 μV/m, which is equivalent to 1.0 x 10^-6 V/m, and the voltage (V) is 1.5 V.

Rearrange the formula to solve for the distance (d) between the plates:

d = \frac{V}{E}

Substitute the known values into the rearranged formula:

d = \frac{1.5}{1.0 	imes 10^{-6}}

Calculate the distance (d) using the substituted values to find how far apart the plates need to be to produce the electric field of 1.0 μV/m with a 1.5 V battery.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Electric Field

An electric field is a region around a charged object where other charged objects experience a force. It is represented by the symbol E and measured in volts per meter (V/m). The strength of the electric field is determined by the voltage applied and the distance between the plates, following the formula E = V/d.

Voltage

Voltage, or electric potential difference, is the measure of the energy per unit charge available to move electrons between two points. It is measured in volts (V) and is a key factor in determining the electric field between two plates. In this scenario, a 1.5 V AA battery provides the voltage needed to create the electric field.

Distance Between Plates

The distance between two parallel plates affects the strength of the electric field created between them. According to the formula E = V/d, the electric field (E) is inversely proportional to the distance (d). To achieve a weak electric field of 1.0 μV/m, the plates must be placed at a specific distance calculated using the given voltage.

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Related Practice

Textbook Question

How much excess charge must be placed on a copper sphere $25.0$ cm in diameter so that the potential of its center, relative to infinity, is $3.75$ kV? What is the potential of the sphere's surface relative to infinity?

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Textbook Question

A metal sphere with radius $r_{a}$ is supported on an insulating stand at the center of a hollow, metal, spherical shell with radius $r_{b}$ . There is charge $+ q$ on the inner sphere and charge $- q$ on the outer spherical shell. Calculate the potential $V (r)$ for (i) $r < r_{a}$ ; (ii) $r_{a} < r < r_{b}$ ; (iii) $r > r_{b}$ . (Hint: The net potential is the sum of the potentials due to the individual spheres.) Take $V$ to be zero when $r$ is infinite.

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Textbook Question

Two large, parallel conducting plates carrying opposite charges of equal magnitude are separated by $2.20$ cm. If the surface charge density for each plate has magnitude $47.0$ nC/m², what is the magnitude of $E$ in the region between the plates?

A metal sphere with radius $r_a$ is supported on an insulating stand at the center of a hollow, metal, spherical shell with radius $r_b$ . There is charge $+q$ on the inner sphere and charge $-q$ on the outer spherical shell. Show that the potential of the inner sphere with respect to the outer is $V_{a} b = q / (4 π ϵ_{0}) (1 / r_{a} - 1 / r_{b})$ .

Two large, parallel conducting plates carrying opposite charges of equal magnitude are separated by $2.20$ cm. What is the potential difference between the two plates?

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Textbook Question

Two large, parallel conducting plates carrying opposite charges of equal magnitude are separated by $2.20$ cm. The surface charge density for each plate has magnitude $47.0$ nC/m^2. If the separation between the plates is doubled while the surface charge density is kept constant at the given value, what happens to the magnitude of the electric field and to the potential difference?

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Certain sharks can detect an electric field as weak as 1.01.0 μμV/m. To grasp how weak this field is, if you wanted to produce it between two parallel metal plates by connecting an ordinary 1.51.5­V AA battery across these plates, how far apart would the plates have to be?

Verified video answer for a similar problem:

Key Concepts

Electric Field

Voltage

Distance Between Plates

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