/*! 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 85 Two \(1.20 \mathrm{~m}\) noncond... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Two \(1.20 \mathrm{~m}\) nonconducting rods meet at a right angle. One rod carries \(+2.50 \mu \mathrm{C}\) of charge distributed uniformly along its length, and the other carries \(-2.50 \mu \mathrm{C}\) distributed uniformly along it (Fig. \(\mathbf{P} 2 \mathbf{1 . 8 5}\) ). (a) Find the magnitude and direction of the electric field these rods produce at point \(P,\) which is \(60.0 \mathrm{~cm}\) from each rod. (b) If an electron is released at \(P\), what are the magnitude and direction of the net force that these rods exert on it?

Short Answer

Expert verified
To provide a short answer, the resultant electric field and force depend on the specific numbers you obtain in your calculations and their directions depend on the positions of the rods. However, since both rods have the same magnitude of charge but opposite signs, the angles will be equal, resulting in a resultant electric field and force along the line connecting the intersection of the rods and point \(P\).

Step by step solution

01

Find the Electric Field Produced by Each Rod

The electric field produced by a uniformly charged rod at a perpendicular distance \(r\) from the rod is given by \(E= \frac{Kq}{r^2}\) where \(K\) is Coulomb's constant (\(8.99 × 10^9 N·m^2/C^2\)), \(q\) is the charge on the rod, and \(r\) is the distance from the rod. Start by calculating the electric field produced by each rod at point \(P\), using the known values \(q = ±2.5μC\), \(K = 8.99 × 10^9 N·m^2/C^2\), and \(r = 60 cm = 0.6 m\). Remember to account for the direction of the electric field: the positive rod will produce a field pointing away from it and the negative rod will produce a field pointing towards it.
02

Find the Net Electric Field

Since the two rods meet at a right angle, their electric fields at point \(P\) will also be at right angles to each other. Therefore, use the Pythagorean theorem to find the magnitude of the resultant electric field \(E_{net} = √(E_1^2 + E_2^2)\). To find the direction of the net electric field, use the tangent function with the electric field magnitudes as opposite and adjacent sides of a right triangle.
03

Calculate the Force on the Electron

Once we have the magnitude and direction of the electric field at \(P\), calculate the force experienced by an electron at that point. The force \(F\) that a charge \(q\) experiences in an electric field \(E\) is given by \(F = qE\). Substitute \(q = -e = -1.6 × 10^-19 C\) (charge of electron) and the found electric field into this formula to find the magnitude of the force. Since the electron is negatively charged, the force it experiences will be in the direction opposite to the electric field.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Coulomb's Law
Coulomb's Law is the cornerstone of understanding electric forces between charged objects. It states that the electric force (\( F \)) between two charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them. Mathematically, it is expressed as:

\[F = \frac{K \cdot |q_1 \cdot q_2|}{r^2}\]where \( K = 8.99 \times 10^9 \, \text{N}\cdot\text{m}^2/\text{C}^2 \) is Coulomb’s constant, \( q_1 \) and \( q_2 \) are the charges, and \( r \) is the distance between them.

This law helps explain how charged rods can exert forces on each other and on other charges nearby. By calculating the electric fields they create, we can further explore how these forces manifest.
Electric Force
The electric force is crucial in determining how charged particles interact with each other. It acts through the electric fields generated by charges and influences any nearby charged objects.

For an electron placed at point \( P \), the force it experiences can be determined using the formula:
  • Formula: \( F = qE \)
  • Where: \( q \) is the charge of the electron (\( -1.6 \times 10^{-19} \, \text{C} \)) and \( E \) is the electric field at point \( P \).
Considering the direction, since an electron has a negative charge, the force is directed opposite to the electric field. Understanding this interaction makes it easier to predict the behavior of charged particles under various conditions.
Charged Rods
Charged rods, when aligned perpendicularly, create unique electric fields at specific points around them. Each rod has a uniform charge distribution along its length, affecting how the field behaves:
  • Positive Rod: Generates an electric field that points away from the rod.
  • Negative Rod: Creates a field pointing towards itself.
At point \( P \), which is equidistant from both rods, these fields interact. Calculating the electric field from each rod involves considering its charge (\( q = \pm 2.5 \, \mu\text{C} \)) and the distance from \( P \), providing insights into their combined effect.
Electric Field Direction
The direction of an electric field is determined by the charge distribution and orientation of the sources. For our rods:
  • The positive rod's field points radially outward.
  • The negative rod's field points inward.
At point \( P \), these directions are perpendicular due to the rods' placement at right angles. The resultant electric field's magnitude can be found using the Pythagorean theorem:

\[E_{\text{net}} = \sqrt{E_1^2 + E_2^2}\]To find the net direction, we use the tangent function, where the two component fields form the sides of a right triangle, providing a clear picture of how they combine at \( P \).

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A proton is projected into a uniform electric field that points vertically upward and has magnitude \(E\). The initial velocity of the proton has a magnitude \(v_{0}\) and is directed at an angle \(\alpha\) below the horizontal. (a) Find the maximum distance \(h_{\max }\) that the proton descends vertically below its initial elevation. Ignore gravitational forces. (b) After what horizontal distance \(d\) does the proton return to its original elevation? (c) Sketch the trajectory of the proton. (d) Find the numerical values of \(h_{\max }\) and \(d\) if \(E=500 \mathrm{~N} / \mathrm{C}, v_{0}=4.00 \times 10^{5} \mathrm{~m} / \mathrm{s},\) and \(\alpha=30.0^{\circ}\)

In a follow-up experiment, a charge of \(+40 \mathrm{pC}\) was placed at the center of an artificial flower at the end of a \(30-\mathrm{cm}\) -long stem. Bees were observed to approach no closer than \(15 \mathrm{~cm}\) from the center of this flower before they flew away. This observation suggests that the smallest external electric field to which bees may be sensitive is closest to which of these values? (a) \(2.4 \mathrm{~N} / \mathrm{C} ;\) (b) \(16 \mathrm{~N} / \mathrm{C} ;\) (c) \(2.7 \times 10^{-10} \mathrm{~N} / \mathrm{C}\) (d) \(4.8 \times 10^{-10} \mathrm{~N} / \mathrm{C}\).

A very long line of charge with charge per unit length \(+8.00 \mu \mathrm{C} / \mathrm{m}\) is on the \(x\) -axis and its midpoint is at \(x=0 .\) A second very long line of charge with charge per length \(-4.00 \mu \mathrm{C} / \mathrm{m}\) is parallel to the \(x\) -axis at \(y=10.0 \mathrm{~cm}\) and its midpoint is also at \(x=0 .\) At what point on the \(y\) -axis is the resultant electric field of the two lines of charge equal to zero?

Point charge \(q_{1}=-5.00 \mathrm{nC}\) is at the origin and point charge \(q_{2}=+3.00 \mathrm{nC}\) is on the \(x\) -axis at \(x=3.00 \mathrm{~cm} .\) Point \(P\) is on the \(y\) -axis at \(y=4.00 \mathrm{~cm} .\) (a) Calculate the electric fields \(\vec{E}_{1}\) and \(\vec{E}_{2}\) at point \(P\) due to the charges \(q_{1}\) and \(q_{2}\). Express your results in terms of unit vectors (see Example 21.6 ). (b) Use the results of part (a) to obtain the resultant field at \(P\), expressed in unit vector form.

A small \(12.3 \mathrm{~g}\) plastic ball is tied to a very light \(28.6 \mathrm{~cm}\) string that is attached to the vertical wall of a room (Fig. \(\mathbf{P} 2 \mathbf{1 . 6 3}\) ). A uniform horizontal electric field exists in this room. When the ball has been given an excess charge of \(-1.11 \mu \mathrm{C},\) you observe that it remains suspended, with the string making an angle of \(17.4^{\circ}\) with the wall. Find the magnitude and direction of the electric field in the room.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.