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Point charge \(q_{1}=-5.00 \mathrm{nC}\) is on the \(x\) -axis at \(x=-0.400 \mathrm{~m}\). Point \(P\) is on the \(x\) -axis at \(x=+0.200 \mathrm{~m}\). Point charge \(q_{2}\) is at the origin. What are the sign and magnitude of \(q_{2}\) if the resultant electric field at point \(P\) is zero?

Short Answer

Expert verified
The sign of \(q_{2}\) needs to be positive to cancel out the rightward electric field created by \(q_{1}\). The magnitude of \(q_{2}\) can be calculated by applying Coulomb's Law, and equating the electric field at point \(P\) due to \(q_{1}\) with the electric field at point \(P\) due to \(q_{2}\).

Step by step solution

01

Calculate the Electric Field Created by \(q_{1}\) at Point \(P\)

We can apply Coulomb's Law to find the electric field produced by \(q_{1}\) at point \(P\). This can be calculated using the formula: \(E_{1} = k \cdot |\frac{q_{1}}{r_{1}^{2}}|\) where \(E_{1}\) is the electric field produced by \(q_{1}\), \(k\) is Coulomb’s constant \(9 \times 10^{9} N.m^{2}/C^{2}\), \(q_{1}\) is the charge at position \(x_1\) and \(r_{1}\) = \(0.60m\) is the distance from \(q_{1}\) to \(P\).
02

Determine the Direction of \(E_{1}\)

The direction of the electric field produced by \(q_{1}\) at point \(P\) would be to the right, due to the negative value of \(q_{1}\). Electric fields point towards negative charges.
03

Calculate the Required Electric Field Created by \(q_{2}\) at Point \(P\)

We want the net electric field to be zero at point \(P\), so the electric field created by \(q_{2}\) at \(P\) needs to cancel out \(E_{1}\). This implies \(E_{2} = E_{1}\), where \(E_{2}\) is the electric field produced by \(q_{2}\) at point \(P\). We can use Coulomb's Law again to calculate \(q_{2}\): \(E_{2} = k \cdot |\frac{q_{2}}{r_{2}^{2}}|\), where \(r_{2} = 0.20m\) is the distance from \(q_{2}\) to \(P\).
04

Determine the Sign of \(q_{2}\)

The electric field created by \(q_{2}\) needs to point to the left to cancel out \(E_{1}\), which points to the right. Only a positive charge \(q_{2}\) placed at the origin can produce an electric field pointing leftwards.

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Key Concepts

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

Coulomb's Law
Let's delve into the foundation of electrostatic interactions: Coulomb's Law. This fundamental principle posits that the force 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. Mathematically, it is expressed as:

\[F = k \cdot \frac{|q_1 \cdot q_2|}{r^2}\]
where:\
    \
  • \(F\) is the magnitude of the electrostatic force,\
  • \
  • \(k\) is Coulomb's constant (approximately \(9 \times 10^9 \mathrm{N \cdot m^2/C^2}\)),\
  • \
  • \(q_1\) and \(q_2\) are the amounts of each charge, and\
  • \
  • \(r\) is the distance between the charges.\
  • \

When dealing with point charges, the force acts along the line joining them, being attractive if the charges are of opposite sign, and repulsive otherwise. In the exercise, we utilize this law to calculate the electric field, which is essentially the force per unit charge, thus shaping our understanding of electrostatic interactions at Point P.
Electric Field
Continuing with our exploration, the concept of an electric field offers a means to understand how a charge affects the space around it. The electric field at a point in space is defined as the force that a positive test charge would experience per unit charge placed at that point. It is vectored, providing us with both the magnitude and the direction of the force a charge would experience. In formula terms:

\[E = k \cdot \frac{|q|}{r^2}\]
Here, \(E\) represents the electric field, and the other symbols have their usual meanings as in Coulomb's Law. Importantly, the electric field direction is radially away from a positive charge and towards a negative charge, which directly informs the second step of the example problem — determining that the electric field created by a negative point charge \(q_1\) at point P would point towards \(q_1\).
Point Charges
The term 'point charges' refers to the idealized version of charges that are so small they can be considered as single points in space. This abstraction is practical for calculations as it allows us to ignore the size of the charges and focus on the distance between their centers. In our exercise, both \(q_1\) and \(q_2\) are treated as point charges, illustrating the effects of charge and distance on the electric field without the complications of charge distribution. It's important to remember the characteristics of point charges to correctly apply Coulomb's Law and understand their role in creating electric fields.
Superposition Principle
Electrostatics is not merely about isolated charges. Often, we encounter multiple charges in a system, necessitating an understanding of the superposition principle. This principle states that the total electric field created by multiple charges is the vector sum of the electric fields created by each individual charge. In other words, electric fields obey the principle of superposition and can be added together to find the net electric field. In the context of our example, we're applying this principle to dictate that the electric field from \(q_2\) must be exactly equal in magnitude but opposite in direction to the electric field from \(q_1\) at point P, resulting in a zero net electric field. Understanding this principle is crucial for solving more complex problems where multiple charges influence a point in space.

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Most popular questions from this chapter

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?

A semicircle of radius \(a\) is in the first and second quadrants, with the center of curvature at the origin. Positive charge \(+Q\) is distributed uniformly around the left half of the semicircle, and negative charge \(-Q\) is distributed uniformly around the right half of the semicircle (Fig. P21.84). What are the magnitude and direction of the net electric field at the origin produced by this distribution of charge?

A point charge is placed at each corner of a square with side length a. All charges have magnitude \(q\). Two of the charges are positive and two are negative (Fig. E21.38). What is the direction of the net electric field at the center of the square due to the four charges, and what is its magnitude in terms of \(q\) and \(a\) ?

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?

A small sphere with positive charge \(q\) and mass \(m\) is released from rest in a uniform electric field \(\vec{E}\) that is directed vertically upward. The magnitude of the field is large enough for the sphere to travel upward when it is released. How long does it take the sphere to travel upward a distance \(d\) after being released from rest? Give your answer in terms of \(q, m, d, E,\) and the acceleration due to gravity, \(g\).

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