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Four identical point charges \((+1.61 \mathrm{nC})\) are placed at the corners of a rectangle, which measures \(3.00 \mathrm{~m}\) by \(5.00 \mathrm{~m} .\) If the electric potential is taken to be zero at infinity, what is the potential at the geometric center of this rectangle?

Short Answer

Expert verified
Answer: The electric potential at the geometric center of the rectangle is \(29.38 \times 10^6 \mathrm{V}\).

Step by step solution

01

Identify the electric potential formula

We'll use the electric potential formula due to a point charge to calculate the electric potential at the center of the rectangle. The formula is: \(V=k_e\frac{q}{r}\), where \(V\) is the electric potential, \(k_e\) is the Coulomb's constant (\(8.99 \times 10^9 \mathrm{Nm^2/C^2}\)), \(q\) is the point charge, and \(r\) is the distance from the charge to the point of interest.
02

Calculate distance for each charge to the center

We'll now find the distances from each corner to the geometric center of the rectangle. Since the opposite corners have equal distances, we have two unique distances to calculate. Let the length \(3.00 \mathrm{~m}\) be along the x-axis and \(5.00 \mathrm{~m}\) along the y-axis from the center. For the two corners on the same horizontal level as the center, the distance \(r_1\) is \(\frac{1}{2} \times 3.00 \mathrm{~m}=1.50 \mathrm{~m}\). For the other two corners, we'll use the Pythagorean theorem to find \(r_2\): $$r_2 = \sqrt{\left(\frac{1}{2} \times 3.00 \mathrm{~m}\right)^2+\left(\frac{1}{2} \times 5.00 \mathrm{~m}\right)^2} = \sqrt{(1.50\mathrm{~m})^2+(2.50\mathrm{~m})^2} = \sqrt{8.25 \mathrm{m^2}} = 2.87 \mathrm{~m}.$$
03

Calculate the electric potential for each charge

Now that we have the distances, we will calculate the electric potential contribution from each charge using the electric potential formula. For the two charges with distance \(r_1\), the electric potential \(V_1=k_e\frac{q}{r_1}=8.99 \times 10^9 \mathrm{Nm^2/C^2} \times \frac{1.61 \times 10^{-9} \mathrm{C}}{1.50 \mathrm{m}}=9.63 \times 10^6 \mathrm{V}\). For the other two charges with distance \(r_2\), the electric potential \(V_2=k_e\frac{q}{r_2}=8.99 \times 10^9 \mathrm{Nm^2/C^2} \times \frac{1.61 \times 10^{-9} \mathrm{C}}{2.87 \mathrm{m}}=5.06 \times 10^6 \mathrm{V}\).
04

Add up the electric potentials

Finally, we will add the electric potentials from all four charges to get the total electric potential at the center of the rectangle. Since each charge is the same, we can multiply the electric potential of distance \(r_1\) by 2, and the electric potential of distance \(r_2\) by 2. The total electric potential \(V_\text{total}\) is: $$V_\text{total}= 2\cdot V_1 + 2 \cdot V_2 = 2 \cdot (9.63 \times 10^6 \mathrm{V}) + 2 \cdot (5.06 \times 10^6 \mathrm{V})= 29.38 \times 10^6 \mathrm{V}.$$ Therefore, the electric potential at the geometric center of the rectangle is \(29.38 \times 10^6 \mathrm{V}\).

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Key Concepts

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

Coulomb's Law
Understanding Coulomb's Law is essential in studying electricity and magnetism. It describes the electrostatic interaction between electrically charged particles. In simple terms, it states that the force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. The formula is expressed as:\[\begin{equation}F = k_e \frac{|q_1 q_2|}{r^2}\end{equation}\]where:
  • F is the magnitude of the force between the charges,
  • k_e is Coulomb's constant (8.99 \times 10^{9} \mathrm{Nm^2/C^2}),
  • q_1 and q_2 are the amounts of the charges,
  • r is the distance between the centers of the two charges.
Coulomb's Law provides the foundational understanding to solve problems related to the forces between charges, which directly translates to computing electric fields and potentials, as seen in the given exercise. When dealing with a point charge, the electric potential due to that charge can be derived from the principles of Coulomb's Law.
Point Charge
The concept of a point charge is a simplification used in physics to model charges with a very small size compared to other distances involved in the problem. A point charge is essentially a charged object whose size can be ignored so the charge can be treated as if it were 'concentrated' at a single point in space. This model is extremely useful in calculations as it allows the use of Coulomb's Law to determine the electric fields and potentials around the charge.In the scenario from the textbook exercise, the four identical charges placed at the corners of a rectangle are all considered point charges. This assumption allows us to use formulas derived from Coulomb's law to calculate the electric potential at the geometric center of the rectangle. The simplification to point charges is valid as long as the distance from the point of interest (the center in this case) is significantly greater than the size of the charges themselves.
Electric Potential Formula
The electric potential at a point in space is a scalar quantity that represents the electric potential energy per unit charge at that point. In practice, it provides a measure of the potential energy a charged particle would have in an electric field, without the complexity of dealing with forces and vectors. The formula for the electric potential due to a point charge is:\[\begin{equation}V = k_e \frac{q}{r}\end{equation}\]where:
  • V is the electric potential, measured in volts (V),
  • k_e is Coulomb's constant,
  • q is the charge of the point charge, and
  • r is the distance from the charge to the point where the potential is being evaluated.
In the context of the textbook exercise, the electric potential formula helps to calculate the potential due to each of the four point charges at the center of the rectangle. Since the electric potential is a scalar quantity, we can simply add the potentials due to each charge to find the total electric potential at the center. This principle allows for a relatively straightforward calculation, even when multiple charges are present, and is an excellent demonstration of the power and utility of the electric potential formula in understanding and solving electrostatic problems.

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Most popular questions from this chapter

Use \(V=\frac{k q}{r}, E_{x}=-\frac{\partial V}{\partial x}, E_{y}=-\frac{\partial V}{\partial y},\) and \(E_{z}=-\frac{\partial V}{\partial z}\) to derive the expression for the electric field of a point charge, \(q\).

A Van de Graaff generator has a spherical conductor with a radius of \(25.0 \mathrm{~cm} .\) It can produce a maximum electric field of \(2.00 \cdot 10^{6} \mathrm{~V} / \mathrm{m}\). What are the maximum voltage and charge that it can hold?

A negatively charged particle revolves in a clockwise direction around a positively charged sphere. The work done on the negatively charged particle by the electric field of the sphere is a) positive. b) negative. c) zero.

Using Gauss's Law and the relation between electric potential and electric field, show that the potential outside a uniformly charged sphere is identical to the potential of a point charge placed at the center of the sphere and equal to the total charge of the sphere. What is the potential at the surface of the sphere? How does the potential change if the charge distribution is not uniform but has spherical (radial) symmetry?

A solid conducting sphere of radius \(R_{1}\) has a charge of \(Q=4.263 \mu C\) evenly distributed over its surface. A second solid conducting sphere of radius \(R_{2}=0.6239 \mathrm{~m}\) is initially uncharged and at a distance of \(10.00 \mathrm{~m}\) from the first sphere. The two spheres are momentarily connected with a wire, which is then removed. The resulting charge on the second sphere is \(1.162 \mu C .\) What is the radius of the first sphere?

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