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Solve the given maximum and minimum problems. The potential energy \(E\) of an electric charge \(q\) due to another charge \(q_{1}\) at a distance of \(r_{1}\) is proportional to \(q_{1}\) and inversely proportional to \(r_{1}\). If charge \(q\) is placed directly between two charges of \(2.00 \mathrm{nC}\) and \(1.00 \mathrm{nC}\) that are separated by \(10.0 \mathrm{mm}\) find the point at which the total potential energy (the sum due to the other two charges) of \(q\) is a minimum.

Short Answer

Expert verified
The charge should be placed at approximately 2.93 mm from the 2.00 nC charge for minimal potential energy.

Step by step solution

01

Understand Potential Energy

The potential energy \( E \) due to a charge \( q_1 \) at distance \( r_1 \) is given by \( E = k \frac{q_1 q}{r_1} \), where \( k \) is a constant. For problems like this, \( k \) often resolves to \( \frac{1}{4\pi \varepsilon_0} \), but since we seek a minimum, the constant \( k \) will not affect the relative distances at which this minimum occurs.
02

Set Up the Equation for Total Potential Energy

Let the charge \( q \) be placed at distance \( x \) from the 2.00 nC charge. Then, it is at \( 10 - x \) mm from the 1.00 nC charge. The total potential energy \( U \) is given by:\[ U = k \left( \frac{2.00q}{x} + \frac{1.00q}{10 - x} \right) \]We want to find \( x \) such that \( U \) is minimized.
03

Differentiate the Potential Energy

Differentiate the expression for total potential energy with respect to \( x \):\[ \frac{dU}{dx} = k \left( \frac{-2.00q}{x^2} + \frac{1.00q}{(10-x)^2} \right) \]After differentiating, set the derivative equal to zero to find critical points.
04

Solve the Equation

Setting the derivative \( \frac{dU}{dx} = 0 \), we have:\[ \frac{-2.00q}{x^2} + \frac{1.00q}{(10-x)^2} = 0 \]This simplifies to:\[ \frac{2}{x^2} = \frac{1}{(10-x)^2} \]Cross-multiply to get:\[ 2(10-x)^2 = x^2 \]Solve the quadratic equation to find \( x \).
05

Solve the Quadratic Equation

Expanding and solving the quadratic:\[ 2(100 - 20x + x^2) = x^2 \]\[ 200 - 40x + 2x^2 = x^2 \]\[ x^2 - 40x + 200 = 0 \]Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = -40, c = 200 \):\[ x = \frac{40 \pm \sqrt{1600 - 800}}{2} \]\[ x = \frac{40 \pm \sqrt{800}}{2} \]\[ x = 5 \pm 5\sqrt{2} \]
06

Determine Feasible Solution

Since \( x \) must be between 0 and 10 mm, \( x = 5 + 5\sqrt{2} \) is not feasible because \( 5\sqrt{2} \approx 7.07 \) adds up to more than 10. The feasible solution is \( x = 5 - 5\sqrt{2} \approx 2.93 \) mm.

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Key Concepts

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

Potential Energy
Potential energy in the context of electric charges is a fundamental notion in physics. It describes the energy a charge possesses due to its position relative to other charges.
When a charge, say charge \( q \), is in the vicinity of another charge \( q_1 \), at a distance \( r_1 \), its potential energy \( E \) can be expressed as \( E = k \frac{q_1 q}{r_1} \). Here, \( k \) is a constant, which often equals \( \frac{1}{4\pi \varepsilon_0} \). However, when looking for a minimum or maximum energy, this constant doesn't influence the result as it is the same throughout the scenario.
The important thing to remember is how potential energy depends inversely on the distance. The farther the charges are apart, the lower the potential energy, and vice versa. This concept is key when trying to optimize energy configurations, like placing a charge \( q \) at a specific point along a line between two other charges.
Differentiation
Differentiation is a crucial tool in calculus, particularly in optimization problems. It allows us to find rates of change and identify points of minimum or maximum values for a given function.
To explore how potential energy changes as you move a charge between two points, you use differentiation. By differentiating the expression for total potential energy with respect to the position \( x \), you can see how the energy changes. You end up with the following expression: \[\frac{dU}{dx} = k \left( \frac{-2.00q}{x^2} + \frac{1.00q}{(10-x)^2} \right)\]By setting this derivative equal to zero, you find critical points where potential energy might be minimized or maximized. Solving this equation involves algebraic manipulation and understanding that critical points are where the derivative changes sign, indicating possible minima or maxima.
Quadratic Equations
When solving for critical points, you often encounter quadratic equations. They are equations of the form \( ax^2 + bx + c = 0 \). A solution can be found using the quadratic formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
  • Discriminant: The term \( b^2 - 4ac \) under the square root is known as the discriminant. It tells us about the number and type of roots:
    • If \( b^2 - 4ac > 0 \), there are two distinct real roots.
    • If \( b^2 - 4ac = 0 \), there is one real root (a double root).
    • If \( b^2 - 4ac < 0 \), there are two complex roots.
While solving for the minimum potential energy, you solve a quadratic equation to find valid \( x \) values, giving two potential solutions. You then check them against the problem constraints, such as \( x \) needing to be in a specific range, to determine the feasible solution. This process reveals the point of minimum potential energy for the charge \( q \).

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Most popular questions from this chapter

In Exercises \(43-54,\) sketch the indicated curves by the methods of this section. You may check the graphs by using a calculator. An electric circuit is designed such that the resistance \(R\) (in \(\Omega\) ) is a function of the current \(i\) (in \(\mathrm{mA}\) ) according to \(R=75-18 i^{2}+8 i^{3}-i^{4} .\) Sketch the graph if \(R \geq 0\) and \(i\) can be positive or negative.

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