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Chapter 19: Problem 18

Identical \(+1.8 \mu \mathrm{C}\) charges are fixed to adjacent corners of a square. What charge (magnitude and algebraic sign) should be fixed to one of the empty corners, so that the total electric potential at the remaining empty corner is \(0 \mathrm{~V} ?\)

Short Answer

Expert verified
The charge required is approximately \(-3.07 \mu \mathrm{C}\).

Step by step solution

01

Understand the Problem

We have a square lattice with side length unspecified. Two adjacent corners have identical charges of \(+1.8 \mu \mathrm{C}\). We need to find the charge needed at one of the other empty corners such that the electric potential at the last corner is \(0 \mathrm{~V}\).
02

Write the Expression for Electric Potential

The electric potential \( V \) at a point due to a point charge \( q \) is given by \( V = \frac{kq}{r} \), where \( k \) is Coulomb's constant \( (8.99 \times 10^9 \mathrm{~Nm}^2/\mathrm{C}^2) \) and \( r \) is the distance from the charge to the point. Since we want the total potential at one corner to be zero, the potentials due to each charge must sum to zero.
03

Calculate Distances Within the Square

Considering a square with side \( a \), the diagonal distance is \( \sqrt{2}a \). For two adjacent corners with \(+1.8 \mu \mathrm{C}\) charges, the potential at the empty corner involves one distance \( a \) and another distance putting \( \sqrt{2}a \) from the corner opposite to where we place the unknown charge \( q \).
04

Equate Potential Contributions

Let us denote the side of the square by \( a \). Then the contributions to the potential at the empty corner due to the adjacent \(+1.8 \mu \mathrm{C}\) charges will be summed as follows: \( V_{total} = \frac{k \times (1.8 \times 10^{-6})}{a} + \frac{k \times (1.8 \times 10^{-6})}{\sqrt{2}a} + \frac{kq}{a} = 0 \).
05

Simplify and Solve for Unknown Charge

Simplify and solve the previous equation for \( q \):\[ \frac{1.8k}{a} + \frac{1.8k}{\sqrt{2}a} + \frac{kq}{a} = 0 \]\[ 1.8 + \frac{1.8}{\sqrt{2}} + q = 0 \]\[ q = -1.8 - \frac{1.8}{\sqrt{2}} = -1.8 - 1.27 \approx -3.07 \mu \mathrm{C} \].
06

Conclusion

The charge needed at one of the empty corners to make the electric potential at the remaining corner zero is approximately \(-3.07 \mu \mathrm{C}\). It must be negative to offset the positive potential contributions from the \(+1.8 \mu \mathrm{C}\) charges.

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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 electrostatics, describing how charges interact. It states that the force between two point charges is proportional to the product of their charges and inversely proportional to the square of the distance between them. The law is mathematically expressed as: \[ F = k \frac{|q_1 q_2|}{r^2} \] where:
  • \( F \) is the force between the charges,
  • \( q_1 \) and \( q_2 \) are the amounts of the charges,
  • \( r \) is the distance between the charges,
  • \( k \) is Coulomb's constant (\(8.99 \times 10^9 \text{Nm}^2/\text{C}^2 \)).
Coulomb's Law helps us understand forces in an electric field, making it essential in calculating electric potential in spaces surrounding point charges.
Point Charges
Point charges are idealized charges that are considered to be concentrated at a single point in space. In physics problems like the one given, point charges simplify calculations by allowing us to focus on their electric effects without worrying about their physical size. Here are some key points:
  • Point charges produce electric fields radiating outward or inward, depending on whether they are positive or negative, respectively.
  • The potential due to a point charge at a certain distance can be calculated using the formula: \[ V = \frac{kq}{r} \]where \( V \) is the electric potential, \( k \) is Coulomb's constant, \( q \) is the amount of the charge, and \( r \) is the distance from the charge.
  • In the problem, the charges are fixed at the corners of a square, which helps in applying electric potential formulas smoothly.
Point charges allow for the calculation of potential and forces easily, making them a crucial concept in dealing with electric fields.
Electric Fields
Electric fields are invisible regions around charges where forces are exerted on other charges. Every point charge creates an electric field that can interact with other charges. Understanding electric fields involves the following:
  • The vector nature: Electric fields have both magnitude and direction, pointing from positive to negative charges.
  • The field due to a point charge is given by: \[ E = \frac{kq}{r^2} \]where \( E \) is the electric field strength, \( q \) is the point charge, \( r \) is the distance from the charge, and \( k \) is Coulomb's constant.
  • These fields can superpose, meaning the total field at a point is the vector sum of fields from individual charges.
Understanding electric fields helps in calculating the forces experienced by charges and also aids in understanding electric potential.
Electrostatics
Electrostatics is the branch of physics that studies electric charges at rest. It encompasses concepts like electric forces, fields, and potentials that are crucial for understanding the problem. Key aspects include:
  • Static electric forces: These are the forces experienced by charges that are stationary.
  • The concept of electric potential energy, which is dependent on the positions of charges and tells us about the work done in moving a charge within an electric field.
  • In electrostatics, like charges repel each other while opposite charges attract, directly affecting how setups like the square in our problem behave.
Electrostatics provides the framework to investigate how charges and fields interact within systems like the one in the exercise. Understanding these interactions is fundamental in resolving questions about charge distribution and electric potential.

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Most popular questions from this chapter

A charge of \(-3.00 \mu \mathrm{C}\) is fixed in place. From a horizontal distance of \(0.0450 \mathrm{~m}, \mathrm{a}\) particle of mass \(7.20 \times 10^{-3} \mathrm{~kg}\) and charge \(-8.00 \mu \mathrm{C}\) is fired with an initial speed of \(65.0 \mathrm{~m} / \mathrm{s}\) directly toward the fixed charge. How far does the particle travel before its speed is zero?

An axon is the relatively long tail-like part of a neuron, or nerve cell. The outer surface of the axon membrane (dielectric constant \(=5,\) thickness \(=1 \times 10^{-8} \mathrm{~m}\) ) is charged positively, and the inner portion is charged negatively. Thus, the membrane is a kind of capacitor. Assuming that an axon can be treated like a parallel plate capacitor with a plate area of \(5 \times 10^{-6} \mathrm{~m}^{2},\) what is its capacitance?

Go Concept Questions Capacitor A and capacitor B both have the same voltage across their plates. However, the energy of capacitor A can melt \(m\) kilograms of ice at \(0^{\circ} \mathrm{C}\), while the energy of capacitor B can boil away the same amount of water at \(100{ }^{\circ} \mathrm{C}\). (a) Which requires more energy, melting the ice or boiling the water? (b) Which capacitor has the greater capacitance? Explain your answers. The capacitance of capacitor \(A\) is \(9.3 \mu F .\) What is the capacitance of capacitor B? Be sure that your answer is consistent with your answers to the Concept Questions.

Four identical charges \((+2.0 \mu C\) each \()\) are brought from infinity and fixed to a straight line. The charges are located \(0.40 \mathrm{~m}\) apart. Determine the electric potential energy of this group.

The inner and outer surfaces of a cell membrane carry a negative and positive charge, respectively. Because of these charges, a potential difference of about \(0.070 \mathrm{~V}\) exists across the membrane. The thickness of the cell membrane is \(8.0 \times 10^{-9} \mathrm{~m}\). What is the magnitude of the electric field in the membrane?
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