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Chapter 25: Problem 42

Two large parallel metal plates are \(1.5 \mathrm{~cm}\) apart and have equal but opposite charges on their facing surfaces. Take the potential of the negative plate to be zero. If the potential halfway between the plates is then \(+5.0 \mathrm{~V}\), what is the electric field in the region between the plates?

Short Answer

Expert verified
The electric field between the plates is \( 666.67 \mathrm{V/m} \).

Step by step solution

01

- Define the Known Values

The distance between the plates is given as \( d = 1.5 \mathrm{cm} = 0.015 \mathrm{m} \). The potential halfway between the plates is given as \( V_{\text{half}} = +5.0 \mathrm{V} \). The potential of the negative plate is defined as zero.
02

- Determine the Potential Difference Between Plates

Since the potential halfway between is \( +5.0 \mathrm{V} \), the total potential difference between the plates \( V_{\text{total}} \) will be \( V_{\text{total}} = 2 \times V_{\text{half}} = 2 \times 5.0 \mathrm{V} = 10.0 \mathrm{V} \).
03

- Use the Relationship Between Electric Field and Potential Difference

The electric field \( E \) between parallel plates is related to the potential difference \( V \) and the distance \( d \) by the formula: \[ E = \frac{V_{\text{total}}}{d} \]
04

- Calculate the Electric Field

Using the formula \[ E = \frac{10.0 \mathrm{V}}{0.015 \mathrm{m}} = 666.67 \mathrm{V/m} \].

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Key Concepts

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

Potential Difference
Understanding the concept of potential difference is crucial in solving problems involving electric fields between parallel plates. The potential difference, also called voltage, is the difference in electric potential between two points.
In the given exercise, the potential halfway between the plates is +5.0 V, and the potential of the negative plate is zero.
This means that the potential difference between the plates is twice the halfway potential because the potential linearly changes between the plates, making it 10.0 V.
The potential difference allows us to understand the energy per unit charge required to move a charge between two points in an electric field.
In this scenario, this difference helps calculate the electric field using the appropriate equation.
Electric Field Equation
The electric field between two parallel plates can be determined using the relationship between the electric field (E), the potential difference (V), and the distance (d) between the plates.
The formula is given by:
This formula tells us that the electric field is a measure of how quickly the potential changes over a given distance. Because the potential difference and the distance between plates are known, we can use this formula to find the electric field.
In our exercise, with a potential difference of 10.0 V and a distance of 0.015 m, the electric field is calculated as: 666.67 V/m This indicates that for every meter in the space between the plates, the potential changes by 666.67 volts.
Distance Between Plates
The distance between the parallel plates plays a crucial role in calculating the electric field.
This distance is often a small value measured in centimeters or millimeters because it impacts how strong the electric field is.
In this exercise, the plates are 1.5 cm apart, which we convert to meters (0.015 m), since it鈥檚 standard to use meters in physics calculations.
A smaller distance between plates leads to a stronger electric field if the potential difference remains constant.
Therefore, knowing the distance accurately is essential for solving such problems using the electric field equation.

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Most popular questions from this chapter

Two tiny metal spheres \(A\) and \(B\) of mass \(m_{A}=5.00 \mathrm{~g}\) and \(m_{B}=10.0 \mathrm{~g}\) have equal positive charges \(q=5.00 \mu \mathrm{C}\). The spheres are connected by a massless nonconducting string of length \(d=1.00 \mathrm{~m}\), which is much greater than the radii of the spheres. (a) What is the electric potential energy of the system? (b) Suppose you cut the string. At that instant, what is the acceleration of each sphere? (c) A long time after you cut the string, what is the speed of each sphere?

Calculate (a) the electric potential established by the nucleus of a hydrogen atom at the average distance \(\left(r=5.29 \times 10^{-11} \mathrm{~m}\right)\) of the atom's electron (take \(V=0\) at infinite distance), (b) the electric potential energy of the atom when the electron is at this radius, and (c) the kinetic energy of the electron, assuming it to be moving in a circular orbit of this radius centered on the nucleus. (d) How much energy is required to ionize the hydrogen atom (that is, to remove the electron from the nucleus so that the separation is effectively infinite)? Express all energies in electron-volts.

An empty hollow metal sphere has a potential of \(+400 \mathrm{~V}\) with respect to ground (defined to be at \(V=0\) ) and has a charge of \(5.0 \times 10^{-9} \mathrm{C}\). Find the electric potential at the center of the sphere.

An infinite non-conducting sheet has a surface charge density \(\sigma=0.10 \mu \mathrm{C} / \mathrm{m}^{2}\) on one side. How far apart are equipotential surfaces whose potentials differ by \(50 \mathrm{~V}\) ?

A particular \(12 \mathrm{~V}\) car battery can send a total charge of \(3.0 \times 10^{5} \mathrm{C}\) through a circuit, from one terminal to the other. (a) How many coulombs of charge does this represent? (b) If this entire charge undergoes a potential difference of \(12 \mathrm{~V}\), how much energy is involved?
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