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Chapter 16: Problem 46

Two point charges, \(q_{1}=+20.0 \mathrm{nC}\) and \(q_{2}=+10.0 \mathrm{nC},\) are located on the \(x\) -axis at \(x=0\) and \(x=1.00 \mathrm{m},\) respectively. Where on the \(x\) -axis is the electric field equal to zero?

Short Answer

Expert verified
Answer: The electric field will be equal to zero at a point approximately \(0.586\,\text{m}\) from the charge \(q_1\) along the x-axis.

Step by step solution

01

1. Set up the problem

First, we need to calculate the electric field created by each charge at a point along the x-axis. The formula for a point charge is: \(E_{point} = \frac{kQ}{r^2}\) where \(E_{point}\) is the electric field at a point, \(k\) is the electrostatic constant \((k \approx 8.99 \times 10^9 Nm^2/C^2)\), \(Q\) is the charge, and \(r\) is the distance from the charge to the point. We will treat the electric field from \(q_{1}\) as positive and the field from \(q_{2}\) as negative, since they are in opposite directions between the charges. Let's call \(x\) as the distance from the first charge \(q_1\) to the point where the electric field is zero.
02

2. Define the electric fields

The electric fields from \(q_1\) and \(q_2\) are: \(E_{1} = \frac{kq_{1}}{x^2}\) \(E_{2} = \frac{kq_{2}}{(1-x)^2}\)
03

3. Set the equations equal and solve for x

Now, we will set the magnitudes of the electric fields equal to each other to find the location x where the net electric field is zero: \(\frac{kq_{1}}{x^2} = \frac{kq_{2}}{(1-x)^2}\) We can cancel out the electrostatic constant \(k\) from both sides of the equation: \(\frac{q_{1}}{x^2} = \frac{q_{2}}{(1-x)^2}\) Now, plug in the given values of \(q_{1} = 20.0 \times 10^{-9} C\) and \(q_{2} = 10.0 \times 10^{-9} C\): \(\frac{20.0 \times 10^{-9}}{x^2} = \frac{10.0 \times 10^{-9}}{(1-x)^2}\)
04

4. Solve for x

Multiply both sides by \(x^2\) and \((1-x)^2\), and simplify: \(20.0 \times 10^{-9}(1-x)^2 = 10.0 \times 10^{-9}x^2\) Divide both sides by \(10.0 \times 10^{-9}\): \(2(1-x)^2 = x^2\) Now expand and simplify: \(2(1 - 2x + x^2) = x^2\) \(2 - 4x + 2x^2 = x^2\) \(x^2 - 4x + 2 = 0\)
05

5. Find x using the quadratic formula

We will now use the quadratic formula to solve for x: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) For our equation, \(a = 1\), \(b = -4\), and \(c = 2\). Plugging in these values: \(x = \frac{4 \pm \sqrt{(-4)^2 - 4(1)(2)}}{2(1)}\) \(x = \frac{4 \pm \sqrt{16 - 8}}{2}\) \(x = \frac{4 \pm \sqrt{8}}{2}\) There are two solutions here: \(x = 2 \pm \sqrt{2}\), but since we are looking for a point between the charges located at \(x=0\) and \(x=1\,\text{m}\), we will choose the smaller value for x: \(x = 2 - \sqrt{2} \approx 0.586\,\text{m}\) So, the electric field will be equal to zero at a point \(\approx 0.586\,\text{m}\) from the charge \(q_1\) along the x-axis.

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Most popular questions from this chapter

Sketch the electric field lines near two isolated and equal (a) positive point charges and (b) negative point charges. Include arrowheads to show the field directions.
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