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Point charge \(A\) is on the \(x\) -axis at \(x=-3.00 \mathrm{~cm}\). At \(x=1.00 \mathrm{~cm}\) on the \(x\) -axis its electric field is \(2700 \mathrm{~N} / \mathrm{C}\). Point charge \(B\) is also on the \(x\) -axis, at \(x=5.00 \mathrm{~cm}\). The absolute magnitude of charge \(B\) is twice that of \(A .\) Find the magnitude and direction of the total electric field at the origin if (a) both \(A\) and \(B\) are positive; (b) both are negative; (c) \(A\) is positive and \(B\) is negative; (d) \(A\) is negative and \(B\) is positive.

Short Answer

Expert verified
For (a) both A and B positive - the total electric field at the origin will be approximately 869 N/C, directed left towards negative x. For (b) both negative - the total electric field at the origin will be approximately 869 N/C, directed right towards positive x. For (c) A positive and B negative - it will be approximately -4833 N/C, meaning the electric field is directed left towards negative x. For (d) A negative and B positive - it will be approximately 4833 N/C, Meaning the electric field is directed right towards positive x.

Step by step solution

01

Identify known quantities

The position of charge A is -3 cm on the X-axis. The position of charge B is 5 cm on the X-axis. The magnitude of the electric field created by A at 1 cm is 2700 N/C and the magnitude of charge B is twice that of A. Convert all measurements to meters for standard calculation.
02

Deduce and calculate unknown quantities

Using Coulomb's law, the absolute value of charge A can be calculated using formula \[E = \frac{K*q}{r^2} ⇒ q_A = E * r^2 / K\]. The distance, r, from A to the point where the electric field is given is 4 cm or 0.04 m. Substitute the given electric field(E) 2700N/C, K (Coulomb's constant) 8.99 * 10^9 N.m^2/C^2, into the formula to calculate the absolute value of charge A. Then find the absolute value of charge B, which is twice the absolute value of A.
03

Calculate the Electric Field from each charge at the origin

Calculate the electric field at the origin made by each charge using the formula \[E = K * |q| / r^2\]. The distance from charge A to the origin is 0.03 m, while it's 0.05 m for charge B.
04

Calculate total Electric Field at the origin for each case

Case (a) and (b): Both charges are the same sign, so the electric fields add. Case (c) and (d): Charges are opposite sign, so the electric fields subtract. Recall that for positive charges, the electric field is directed away from the charge, whereas for negative charges, it's towards them.

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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 foundational concept in understanding the interactions between point charges. Simply put, it states that the force (\f\(F\f\)) between two point charges is directly proportional to the product of their charges (\f\(q_1\f\) and \f\(q_2\f\)) and inversely proportional to the square of the distance (\f\(r\f\)) separating them. The equation can be expressed as: \f[ F = K \frac{|q_1 \times q_2|}{r^2} \f], where \f\(K\f\) is Coulomb's constant, approximately equal to \f\(8.99 \times 10^9 N.m^2/C^2\f\).

This fundamental law helps us calculate the magnitude of electrostatic forces and is pivotal for understanding electric fields around point charges. For example, in the given exercise, we use Coulomb's law to deduce the absolute value of charge \f\(A\f\) by rearranging the formula to solve for \f\(q_A\f\), taking into account the known electric field intensity and distance.
Electric Field Intensity
The concept of electric field intensity (\f\(E\f\)) is essential in visualizing and calculating how a charge exerts force on other charges around it without physical contact. It is defined as the force experienced by a unit positive charge placed in the vicinity of another charge. The electric field intensity for a point charge is given by the formula: \f[ E = \frac{K \times |q|}{r^2} \f], where \f\(q\f\) is the charge creating the field, \f\(r\f\) is the distance from the charge to the point in question, and \f\(K\f\) is again Coulomb's constant.

In our exercise, the electric field intensity exerted by charge \f\(A\f\) at a specific point is given, and we extend this concept to calculate the fields at the origin due to both \f\(A\f\) and \f\(B\f\), which helps us determine the total electric field at the origin for various scenarios of charge configurations.
Superposition Principle
The superposition principle plays a crucial role when dealing with multiple charges. It allows us to calculate the net electric field produced by several charges by considering each individual charge's contribution separately and then algebraically adding these vector quantities. According to this principle, the total electric field (\f\(\textbf{E}_{\text{total}}\f\)) created by multiple point charges is the vector sum of the electric fields (\f\(\textbf{E}_{i}\f\)) produced by each charge independently: \f[ \textbf{E}_{\text{total}} = \textbf{E}_1 + \textbf{E}_2 + ... + \textbf{E}_n \f].

This principle is pivotal in solving our exercise since charge \f\(A\f\) and charge \f\(B\f\) both contribute to the electric field at the origin. Depending on whether the charges are of the same or opposite signs, their fields either add or subtract vectorially to produce the net field.
Vector Addition of Electric Fields
The process of adding electric fields vectorially is an application of the superposition principle specifically to vector quantities. Remember that electric fields have both magnitude and direction, so when combining the fields from multiple charges, we must consider the vector character of these fields. For point charges placed along a line (like in our exercise), this simplifies to a one-dimensional problem where the fields either add or subtract according to their direction.

However, for cases where charges are not collinear, one might have to employ vector addition, involving breaking down the fields into components, typically using a Cartesian coordinate system, and then summing these components. To solve the given textbook problem, we perform such an addition (or subtraction) based on the sign and alignment of charges, to find the resulting net field at the origin.

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Most popular questions from this chapter

A uniform line of charge with length \(20.0 \mathrm{~cm}\) is along the \(x\) -axis, with its midpoint at \(x=0 .\) Its charge per length is \(+4.80 \mathrm{nC} / \mathrm{m}\) A small sphere with charge \(-2.00 \mu \mathrm{C}\) is located at \(x=0, y=5.00 \mathrm{~cm}\) What are the magnitude and direction of the force that the charged sphere exerts on the line of charge?

A proton is placed in a uniform electric field of \(2.75 \times 10^{3} \mathrm{~N} / \mathrm{C} .\) Calculate (a) the magnitude of the electric force felt by the proton; (b) the proton's acceleration; (c) the proton's speed after \(1.00 \mu \mathrm{s}\) in the field, assuming it starts from rest.

A nerve signal is transmitted through a neuron when an excess of \(\mathrm{Na}^{+}\) ions suddenly enters the axon, a long cylindrical part of the neuron. Axons are approximately \(10.0 \mu \mathrm{m}\) in diameter, and measurements show that about \(5.6 \times 10^{11} \mathrm{Na}^{+}\) ions per meter (each of charge \(+e\) ) enter during this process. Although the axon is a long cylinder, the charge does not all enter everywhere at the same time. A plausible model would be a series of point charges moving along the axon. Consider a \(0.10 \mathrm{~mm}\) length of the axon and model it as a point charge. (a) If the charge that enters each meter of the axon gets distributed uniformly along it, how many coulombs of charge enter a \(0.10 \mathrm{~mm}\) length of the axon? (b) What electric field (magnitude and direction) does the sudden influx of charge produce at the surface of the body if the axon is \(5.00 \mathrm{~cm}\) below the skin? (c) Certain shark can respond to electric fields as weak as \(1.0 \mu \mathrm{N} / \mathrm{C}\). How far from this segment of axon could a shark be and still detect its electric field?

Consider an infinite flat sheet with positive charge density \(\sigma\) in which a circular hole of radius \(R\) has been cut out. The sheet lies in the \(x y\) -plane with the origin at the center of the hole. The sheet is parallel to the ground, so that the positive \(z\) -axis describes the "upward" direction. If a particle of mass \(m\) and negative charge \(-q\) sits at rest at the center of the hole and is released, the particle, constrained to the \(z\) -axis, begins to fall. As it drops farther beneath the sheet, the upward electric force increases. For a sufficiently low value of \(m,\) the upward electrical attraction eventually exceeds the particle's weight and the particle will slow, come to a stop, and then rise back to its original position. This sequence of events will repeat indefinitely. (a) What is the electric field at a depth \(\Delta\) beneath the origin along the negative \(z\) -axis? (b) What is the maximum mass \(m_{\max }\) that would prevent the particle from falling indefinitely? (c) If \(m

Two small aluminum spheres, each having mass \(0.0250 \mathrm{~kg}\), are separated by \(80.0 \mathrm{~cm}\). (a) How many electrons does each sphere contain? (The atomic mass of aluminum is \(26.982 \mathrm{~g} / \mathrm{mol}\), and its atomic number is \(13 .\) ) (b) How many electrons would have to be removed from one sphere and added to the other to cause an attractive force between the spheres of magnitude \(1.00 \times 10^{4} \mathrm{~N}\) (roughly 1 ton)? Assume that the spheres may be treated as point charges. (c) What fraction of all the electrons in each sphere does this represent?

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