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At a distance of 1.60 \(\mathrm{m}\) from a point charge of \(+2.00 \mu \mathrm{C},\) there is an equipotential surface. At greater distances there additional equipotential surfaces. The potential difference between any two successive surfaces is \(1.00 \times 10^{3} \mathrm{V}\) . Starting at a distance of 1.60 \(\mathrm{m}\) and moving radially outward, how many of the additional equipotential surfaces are crossed by the time the electric field has shrunk to one-half of its initial value? Do not include the starting surface.

Short Answer

Expert verified
3 additional equipotential surfaces are crossed.

Step by step solution

01

Understand Electric Field and Potential

The electric field due to a point charge is given by the expression: \( E = \frac{k \cdot Q}{r^2} \), where \( k \) is Coulomb's constant (approximately \(8.99 \times 10^9 \, \mathrm{N \cdot m^2/C^2} \)), \( Q \) is the charge, and \( r \) is the distance from the charge. The electric potential is given by \( V = \frac{k \cdot Q}{r} \). An equipotential surface is where the potential \( V \) is constant.
02

Calculate Initial Electric Field

At a distance of 1.60 m, calculate the initial electric field using the formula \( E = \frac{k \cdot Q}{r^2} \). With \( Q = 2.00 \times 10^{-6} \mathrm{C} \), \( r = 1.60 \, \mathrm{m} \), and \( k = 8.99 \times 10^9 \, \mathrm{N \cdot m^2/C^2} \), the initial electric field is: \( E = \frac{8.99 \times 10^9 \cdot 2.00 \times 10^{-6}}{(1.60)^2}.\)
03

Determine Half of Initial Electric Field

Calculate half of the initial electric field value obtained in Step 2. This new electric field strength will determine the distance at which the field is halved.
04

Find New Distance Where Field is Halved

Using the relationship for electric fields, \( E_{new} = \frac{E_{initial}}{2} = \frac{k \cdot Q}{r_{new}^2} \), solve for \( r_{new} \). Equating \( \frac{k \cdot Q}{r^2} / 2 = \frac{k \cdot Q}{r_{new}^2} \) simplifies to \( r_{new}^2 = 2r^2 \), so \( r_{new} = \sqrt{2} r \). Thus, \( r_{new} = \sqrt{2} \times 1.60 \approx 2.26 \, \mathrm{m} \).
05

Calculate Number of Crossed Equipotential Surfaces

The potential difference between surfaces is \( 1000 \, \mathrm{V} \). Compute the potential at 1.60 m using \( V_{1.60} = \frac{k \cdot Q}{1.60} \), and at the new distance: \( V_{new} = \frac{k \cdot Q}{r_{new}} \). Determine the number of crossed surfaces: \( n = \frac{V_{1.60} - V_{new}}{1000} \).

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Key Concepts

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

Equipotential Surface
An equipotential surface is a region in space where every point has the same electric potential. This means that no work is required to move a charge along this surface, as the potential energy remains constant.

Equipotential surfaces are always perpendicular to electric field lines. This is because the electric field is a measure of the change in electric potential over a distance. If you move along an equipotential surface, there is no change in potential, and hence, the movement is orthogonal to the field direction.

For a point charge, equipotential surfaces are spherical shells centered around the charge. As you move further away from the charge, these surfaces become larger, signifying that the potential decreases with distance.
Electric Potential
Electric potential, often represented as (V), is the electric potential energy per unit charge at a point in space. It's a measure of how much potential energy a test charge would have at a specific location under the influence of an electric field.

The potential difference between two points is what causes charges to move, somewhat like how a difference in height causes water to flow downhill. The formula for electric potential due to a point charge is given by:
  • \( V = \frac{k \cdot Q}{r} \)
where \( V \) is the potential, \( k \) is Coulomb’s constant, \( Q \) is the charge, and \( r \) is the distance from the charge.

It's important to remember that electric potential is a scalar quantity, which means it doesn’t have a direction. It describes the potential energy landscape in space around charges.
Coulomb's Law
Coulomb’s law describes the force between two point charges. It tells us that the electric force (\( F \) ) between two charges is proportional to the product of the absolute values of the charges and inversely proportional to the square of the distance between them.

The formula is:
  • \( F = \frac{k \cdot |Q_1 \cdot Q_2|}{r^2} \)
where \( F \) is the force, \( Q_1 \) and \( Q_2 \) are the values of the two charges, \( r \) is the distance separating the charges, and \( k \) is Coulomb's constant (\(8.99 \times 10^9 \, \mathrm{N \cdot m^2/C^2}\)).

This fundamental principle not only explains the interaction between charges but also forms the basis for understanding how electric fields behave around point charges, as seen in the problem solution.
Point Charge
A point charge is an idealized model of a charge that is located at a single point in space. It’s used as a simplification to understand electric fields and potentials without considering the size or shape of the charged object.

In physics, point charges are often used to simplify complex systems, where we assume the charge distribution is very small compared to the distances involved in the problem. This allows us to use simpler equations like those derived from Coulomb's law.

Point charges are indispensable in the study of electrostatics because they form the simplest model of how charges interact with each other and create electric fields. The electric field around a point charge decreases in intensity as one moves further away from the charge, behaving inversely with the square of the distance, which is crucial for solving related problems.

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Most popular questions from this chapter

Equipotential surface \(A\) has a potential of \(5650 \mathrm{V},\) while equipotential surface \(B\) has a potential of 7850 \(\mathrm{V}\) . A particle has a mass of \(5.00 \times 10^{-2} \mathrm{kg}\) and a charge of \(+4.00 \times 10^{-5} \mathrm{C}\) . The particle has a speed of 2.00 \(\mathrm{m} / \mathrm{s}\) on surface \(A\) . A nonconservative outside force is applied to the particle, and it moves to surface \(B\) , arriving there with a speed of 3.00 \(\mathrm{m} / \mathrm{s}\) . How much work is done by the outside force in moving the particle from \(A\) to \(B ?\)

Refer to Multiple-Concept Example 3 to review the concepts that are needed here. A cordless electric shaver uses energy at a rate of 4.0 \(\mathrm{W}\) from a rechargeable \(1.5-\mathrm{N}\) battery. Each of the charged particles that the battery delivers to the shaver carries a charge that has a magnitude of \(1.6 \times 10^{-19} \mathrm{C}\) . A fully charged battery allows the shaver to be used for its maximum operation time, during which \(3.0 \times 10^{22}\) of the charged particles pass between the terminals of the battery as the shaver operates. What is the shaver's maximum operation time?

An axon is the relatively long tail-like part of a neuron, or nerve cell. The outer surface of the axon membrane (dielectric constant \(=5,\) thickness \(=1 \times 10^{-8} \mathrm{m}\) ) is charged positively, and the inner portion is charged negatively. Thus, the membrane is a kind of capacitor. Assuming that the membrane acts like a parallel plate capacitor with a plate area of \(5 \times 10^{-6} \mathrm{m}^{2},\) what is its capacitance?

An empty parallel plate capacitor is connected between the terminals of a \(9.0-\mathrm{V}\) battery and charged up. The capacitor is then disconnected from the battery, and the spacing between the capacitor plates is doubled. As a result of this change, what is the new voltage between the plates of the capacitor?

A charge of \(+125 \mu \mathrm{C}\) is fixed at the center of a square that is 0.64 \(\mathrm{m}\) on a side. How much work is done by the electric force as a charge of \(+7.0 \mathrm{\muC}\) is moved from one corner of the square to any other empty corner? Explain.

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