/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 46 A charge of \(-9.0 \mathrm{nC}\)... [FREE SOLUTION] | 91Ó°ÊÓ

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A charge of \(-9.0 \mathrm{nC}\) is uniformly distributed around a thin plastic ring lying in a \(y z\) plane with the ring center at the origin. \(\mathrm{A}-6.0 \mathrm{pC}\) particle is located on the \(x\) axis at \(x=3.0 \mathrm{~m}\). For a ring radius of \(1.5 \mathrm{~m},\) how much work must an external force do on the particle to move it to the origin?

Short Answer

Expert verified
The work required is approximately \(1.79 \times 10^{-10} \text{ J}\).

Step by step solution

01

Identify Constants and Variables

We start by listing the known quantities:- Charge of the ring, \(Q = -9.0 \text{ nC} = -9.0 \times 10^{-9} \text{ C}\)- Charge of the particle, \(q = -6.0 \text{ pC} = -6.0 \times 10^{-12} \text{ C}\)- Distance from origin to initial position of the particle, \(x_i = 3.0 \text{ m}\)- Radius of the ring, \(R = 1.5 \text{ m}\)- Final position of the particle at the origin, \(x_f = 0 \text{ m}\).Now, we must determine how much work is required to move the particle from \(x = 3.0 \text{ m}\) to the origin, \(x = 0 \text{ m}\).
02

Calculate the Initial Electric Potential Energy

The electric potential \(V\) at a point on the axis of a uniformly charged ring is given by:\[ V = \frac{kQ}{\sqrt{x^2 + R^2}} \]where \(k = 8.99 \times 10^9 \text{ N m}^2/\text{C}^2\) is Coulomb's constant.At \(x_i = 3.0 \text{ m}\), the potential \(V_i\) is:\[ V_i = \frac{8.99 \times 10^9 \times (-9.0 \times 10^{-9})}{\sqrt{3.0^2 + 1.5^2}} \]This simplifies to:\[ V_i = \frac{-80.91}{3.3541} \approx -24.12 \text{ V}\]The initial potential energy \(U_i\) when the particle is at \(x_i\) is:\[ U_i = q \cdot V_i = (-6.0 \times 10^{-12}) \times (-24.12) \approx 1.45 \times 10^{-10} \text{ J}\]
03

Calculate the Final Electric Potential Energy

At the origin \(x_f = 0\), the potential \(V_f\), is:\[ V_f = \frac{8.99 \times 10^9 \times (-9.0 \times 10^{-9})}{1.5} \approx -53.94 \text{ V}\]The final potential energy \(U_f\) when the particle is at \(x_f\) is:\[ U_f = q \times V_f = (-6.0 \times 10^{-12}) \times (-53.94) \approx 3.24 \times 10^{-10} \text{ J}\]
04

Calculate the Work Done by External Force

The work done \(W\) by an external force in moving the particle from \(x_i\) to \(x_f\) is the change in potential energy:\[ W = U_f - U_i = 3.24 \times 10^{-10} - 1.45 \times 10^{-10} \]This simplifies to:\[ W = 1.79 \times 10^{-10} \text{ J}\]
05

Conclusion

The work required by an external force to move the particle to the origin is calculated to be approximately \(1.79 \times 10^{-10} \text{ J}\).

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Key Concepts

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

Coulomb's Law
Coulomb's Law is a fundamental principle used to calculate the force between two point charges. It's key to understanding electric interactions. It states that the magnitude of the force (F) between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. The formula for Coulomb's Law is:\[ F = k \frac{|q_1 q_2|}{r^2} \]where:
  • \( k \) is Coulomb's constant (\( 8.99 \times 10^9 \text{ N m}^2/\text{C}^2 \)),
  • \( q_1 \) and \( q_2 \) are the magnitudes of the charges,
  • \( r \) is the distance between the centers of the two charges.
Understanding Coulomb's Law allows us to calculate forces in scenarios where charge distribution is uniform or point-like.
In this exercise, the law helps establish potential energy changes when moving charges, although the force calculation is implicit.
Work Done by External Forces
The concept of work done by an external force is central to understanding energy changes in physics. When a charge is moved in an electric field, work is either done on the field or by the field. Here, the external force does work to move the charge from one point to another against the electric field.
The work done (W) by an external force is calculated by the change in electric potential energy (U). It's derived with:\[ W = U_f - U_i \]where:
  • \( U_f \) is the final potential energy
  • \( U_i \) is the initial potential energy.
The external force must counteract the field's inherent push or pull on the charge.
In this problem, we calculated how much energy is required to move the particle against the electric field created by the charged ring, showing it involves increasing the particle's potential energy from its initial to final state.
Uniform Charge Distribution
Uniform charge distribution refers to a scenario where charges are spread evenly across an object, such as a ring, plane, or sphere. This concept simplifies calculations by assuming the charge distribution does not change with position, which is crucial for simplifying complex electric field and potential calculations.
For a uniformly charged ring, like in this exercise:
  • The charge is symmetrically distributed along the ring's circumference.
  • This symmetry allows the use of simplifications in calculating potential energy and potential difference on points along the axis of the ring.
The electric potential (V) produced by a uniformly charged ring at a point on its axis can be derived using this symmetry and is given by the formula:\[ V = \frac{kQ}{\sqrt{x^2 + R^2}} \]where:
  • \( Q \) is the total charge on the ring.
  • \( x \) is the distance from the center of the ring to the point of interest.
  • \( R \) is the radius of the ring.
Understanding this distribution helps students visualize electric fields around symmetrical objects and the potential energy calculations involved when moving charges within these fields.

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Most popular questions from this chapter

A particular \(12 \mathrm{~V}\) car battery can send a total charge of \(84 \mathrm{~A} \cdot \mathrm{h}\) (ampere-hours) through a circuit, from one terminal to the other. (a) How many coulombs of charge does this represent? (Hint: See Eq. \(21-3 .)\) (b) If this entire charge undergoes a change in electric potential of \(12 \mathrm{~V},\) how much energy is involved?

Two isolated, concentric, conducting spherical shells have radii \(R_{1}=0.500 \mathrm{~m}\) and \(R_{2}=1.00 \mathrm{~m},\) uniform charges \(q_{1}=+2.00 \mu \mathrm{C}\) and \(q_{2}=+1.00 \mu \mathrm{C},\) and negligible thicknesses. What is the magnitude of the electric field \(E\) at radial distance (a) \(r=4.00 \mathrm{~m}\) (b) \(r=0.700 \mathrm{~m},\) and \((\mathrm{c}) r=0.200 \mathrm{~m} ?\) With \(V=0\) at infinity, what is \(V\) at \((\mathrm{d}) r=4.00 \mathrm{~m},(\mathrm{e}) r=1.00 \mathrm{~m},(\mathrm{f}) r=0.700 \mathrm{~m},(\mathrm{~g}) r=0.500 \mathrm{~m}\) (h) \(r=0.200 \mathrm{~m},\) and (i) \(r=0 ?(j)\) Sketch \(E(r)\) and \(V(r)\).

A metal sphere of radius \(15 \mathrm{~cm}\) has a net charge of \(3.0 \times 10^{-8} \mathrm{C}\). (a) What is the electric field at the sphere's surface? (b) If \(V=0\) at infinity, what is the electric potential at the sphere's surface? (c) At what distance from the sphere's surface has the electric potential decreased by \(500 \mathrm{~V} ?\)

The electric field in a region of space has the components \(E_{y}=\) \(E_{z}=0\) and \(E_{x}=(4.00 \mathrm{~N} / \mathrm{C}) x .\) Point \(A\) is on the \(y\) axis at \(y=3.00 \mathrm{~m},\) and point \(B\) is on the \(x\) axis at \(x=4.00 \mathrm{~m}\). What is the potential difference \(V_{B}-V_{A} ?\)

Two charges \(q=+2.0 \mu \mathrm{C}\) are fixed a distance \(d=2.0 \mathrm{~cm}\) apart (Fig. \(24-69\) ). (a) With \(V=0\) at infinity, what is the electric potential at point \(C ?\) (b) You bring a third charge \(q=+2.0 \mu \mathrm{C}\) from infinity to \(C .\) How much work must you do? (c) What is the potential energy \(U\) of the three charge configuration when the third charge is in place?

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