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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

Positive point charges \(q\) and \(2 q\) are located at \(x=0\) and \(x=3 d,\) respectively. What is the electric field at \(x=2 d\) (point \(S\) )?
Positive point charges \(q\) and \(2 q\) are located at \(x=0\) and \(x=3 d,\) respectively. What is the electric field at \(x=d\) (point \(P\) )?
In a cathode ray tube, electrons initially at rest are accelerated by a uniform electric field of magnitude $4.0 \times 10^{5} \mathrm{N} / \mathrm{C}\( during the first \)5.0 \mathrm{cm}$ of the tube's length; then they move at essentially constant velocity another \(45 \mathrm{cm}\) before hitting the screen. (a) Find the speed of the electrons when they hit the screen. (b) How long does it take them to travel the length of the tube?
A hollow conducting sphere of radius \(R\) carries a negative charge \(-q .\) (a) Write expressions for the electric field \(\overrightarrow{\mathbf{E}}\) inside \((rR)\) the sphere. Also indicate the direction of the field. (b) Sketch a graph of the field strength as a function of \(r\). [Hint: See Conceptual Example \(16.8 .]\)

Two small metal spheres are \(25.0 \mathrm{cm}\) apart. The spheres have equal amounts of negative charge and repel each other with a force of \(0.036 \mathrm{N}\). What is the charge on each sphere?
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