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Chapter 24: Problem 41

Two point charges, one \(+400.0 \mathrm{nC}\) and the other \(-400.0 \mathrm{nC}\), located \(20.00 \mathrm{~cm}\) to the right of the first, are in vacuum. Determine the electric field (magnitude and direction) at a point midway between the charges.

Short Answer

Expert verified
The electric field at the midpoint is \(7.192 \times 10^5 \, \mathrm{N/C}\) to the right.

Step by step solution

01

Identify the Problem

We need to find the electric field at the midpoint between two point charges with equal magnitude but opposite signs. The charges are separated by a distance of 20.00 cm, so the midpoint is 10.00 cm from each charge.
02

Understand Electric Field from a Charge

The electric field due to a single point charge is given by the formula \( E = \frac{k |q|}{r^2} \), where \( k \) is Coulomb's constant, \( q \) is the charge, and \( r \) is the distance from the charge.
03

Calculate Electric Field from +400 nC Charge

For the +400 nC charge, \( E_+ = \frac{(8.99 \times 10^9 \, \mathrm{N \, m^2/C^2})(400 \times 10^{-9} \, \mathrm{C})}{(0.1 \, \mathrm{m})^2} = 3.596 \times 10^5 \, \mathrm{N/C} \). The direction is away from the positive charge.
04

Calculate Electric Field from -400 nC Charge

For the -400 nC charge, \( E_- = \frac{(8.99 \times 10^9 \, \mathrm{N \, m^2/C^2})(400 \times 10^{-9} \, \mathrm{C})}{(0.1 \, \mathrm{m})^2} = 3.596 \times 10^5 \, \mathrm{N/C} \). The direction is towards the negative charge.
05

Determine Net Electric Field

Since both electric fields have the same magnitude but different directions, they add up. The electric field due to the positive charge points to the right, and the electric field due to the negative charge also points to the right. Therefore, the net electric field at the midpoint is \( E_{net} = 2 \times 3.596 \times 10^5 \, \mathrm{N/C} = 7.192 \times 10^5 \, \mathrm{N/C} \) directed to the right.

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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 fundamental in understanding the interaction between electric charges. It describes how the electric force between two point charges depends on the magnitude of the charges and the distance between them. The formula for the electric force is
  • \( F = k \frac{|q_1 q_2|}{r^2}\)
where:
  • \(F\) is the magnitude of the force between the charges,
  • \(k\) is Coulomb's constant \((8.99 \times 10^9 \, \mathrm{N \, m^2/C^2})\),
  • \(q_1\) and \(q_2\) are the magnitudes of the charges,
  • \(r\) is the distance between the centers of the two charges.

It's important to note that the force is attractive if the charges are of opposite signs and repulsive if they are of the same sign. Coulomb's law not only helps in calculating this force but also finds applications in determining the electric fields created by point charges.
Point Charges
Point charges are an abstraction used to simplify problems where the size of the charged objects is very small compared to the distances involved. In physics, they are treated as having all their charge concentrated at a single point in space. When dealing with point charges, we usually assume:
  • Charges can be treated as points when the distance between them is much larger than the size of the objects possessing the charges.
  • Analyzing the electric field or force between point charges can predict real-world behavior of charged particles like electrons and protons.

In the given exercise, the two point charges of \(+400.0 \, \mathrm{nC}\) and \(-400.0 \, \mathrm{nC}\) are considered such that their separation distance is the primary variable impacting interactions, as opposed to their physical dimensions.
Electric Field Direction
When dealing with electric fields in the context of charged particles, the field's direction is crucial for understanding how it influences other charges. The direction of the electric field produced by a positive charge is always radially outward from the charge, while for a negative charge, it's radially inward.
For example, at a point between two charges, the direction of the resulting electric field depends on the contributions from each charge:
  • A positive charge creates an electric field pointing away from itself.
  • A negative charge creates an electric field pointing towards itself.

In the exercise above, despite opposite directions of individual electric fields at the midpoint, their net effect results in a single direction. Since both field vectors have the same magnitude in this case, and both add up in the same spatial direction (towards the negative charge and away from the positive one), the net electric field points towards the negative charge, to the right of the midpoint.

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Most popular questions from this chapter

Determine the acceleration of a proton \(\left(q=+e, m=1.67 \times 10^{-27}\right.\) kg) immersed in an electric field of strength \(0.50 \mathrm{kN} / \mathrm{C}\) in vacuum How many times is this acceleration greater than that due to gravity?

Find the ratio of the Coulomb electric force \(F_{E}\) to the gravitational force \(F_{G}\) between two electrons in vacuum. From Coulomb's Law and Newton's Law of gravitation, The electric force is much stronger than the gravitational force. $$ \begin{array}{c} F_{E}=k \frac{q^{2}}{r^{2}} \text { and } F_{G}=G \frac{m^{2}}{r^{2}} \\ \text { Therefore, } \quad \begin{aligned} \frac{F_{E}}{F_{G}} &=\frac{k q_{*}^{2} / r^{2}}{G m^{2} / r^{2}}=\frac{k q_{i}^{2}}{G m^{2}} \\ =& \frac{\left(9.0 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}\right)\left(1.6 \times 10^{-19} \mathrm{C}\right)^{2}}{\left(6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2}\right)\left(9.1 \times 10^{-31} \mathrm{~kg}\right)^{2}}=4.2 \times 10^{42} \end{aligned} \end{array} $$

Two small charged spheres are placed in vacuum on the \(x\) -axis: \(+3.0 \mu \mathrm{C}\) at \(x=0\) and \(-5.0 \mu \mathrm{C}\) at \(x=40 \mathrm{~cm} .\) Where must a third charge \(q\) be placed if the force it experiences is to be zero? The situation is represented in Fig. 24-4. We know that \(q\) must be placed somewhere on the \(x\) -axis. (Why?) Suppose that \(q\) is positive. When it is placed in interval \(B C\), the two forces on it are in the same direction and cannot cancel. When it is placed to the right of \(C\), the attractive force from the \(-5 \mu \mathrm{C}\) charge is always larger than the repulsion of the \(+3.0 \mu \mathrm{C}\) charge. Therefore, the force on \(q\) cannot be zero in this region. Only in the region to the left of \(B\) can cancellation occur. (Can you show that this is also true if \(q\) is negative?) For \(q\) placed as shown, when the net force on it is zero, we have \(F_{E 3}=F_{E 5}\) and so, for distances in meters, $$ k_{0} \frac{q\left(3.0 \times 10^{-6} \mathrm{C}\right)}{d^{2}}=k_{0} \frac{q\left(5.0 \times 10^{-6} \mathrm{C}\right)}{(0.40 \mathrm{~m}+d)^{2}} $$ After canceling \(q, k_{0}\), and \(10^{-6} \mathrm{C}\) from each side, cross- multiply to obtain $$ 5 d^{2}=3.0(0.40+d)^{2} \text { or } d^{2}-1.2 d-0.24=0 $$ Using the quadratic formula, $$ d=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}=\frac{1.2 \pm \sqrt{1.44+0.96}}{2}=0.60 \pm 0.775 \mathrm{~m} $$ Two values, \(1.4 \mathrm{~m}\) and \(-0.18 \mathrm{~m}\), are therefore found for \(d\). The first is the correct one; the second gives the point in \(B C\) where the two forces have the same magnitude but do not cancel.

Determine the force between two free electrons spaced 1.0 angstrom \(\left(10^{-10} \mathrm{~m}\right)\) apart in vacuum.

If two equal point charges, each of \(1 \mathrm{C}\), were separated in air by a distance of \(1 \mathrm{~km}\), what would be the force between them?
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