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Chapter 9: Problem 19

The sign of \(q\) remains positive and the sign of \(2 q\) is changed to negative. Is there any point along the \(x\)-axis where the electric field could be zero? (A) Yes, somewhere to the left of the charge marked \(q\) (B) Yes, somewhere to the right of the charge marked \(2 q\) (C) Yes, between the two charges but closer to \(q\) (D) Yes, between the two charges but closer to \(2 q\) (E) No, the field can never be zero

Short Answer

Expert verified
Answer: Yes, between the two charges but closer to \(2q\).

Step by step solution

01

1. Determine the electric field created by the charges

The electric field created by a charge \(Q\) at a point \(P\) placed at a distance \(r\) is given by the formula $$E = \frac{kQ}{r^2}$$where \(k\) is Coulomb's constant. For charge \(q\), the electric field at point \(P\) is $$E_q = \frac{kq}{r_1^2}$$ and for charge \(2q\), the electric field at point \(P\) is $$E_{2q} = \frac{k(2q)}{r_2^2}$$ where \(r_1\) and \(r_2\) are the distances of point \(P\) from charges \(q\) and \(2q\) respectively. To find a point where the electric field is zero, these two electric fields must have equal magnitudes but opposite directions.
02

2. Set up the equation for equal magnitudes of the electric fields

Since the directions of the electric fields are opposite, we can set up the following equation for equal magnitude:$$\frac{kq}{r_1^2} = \frac{k(2q)}{r_2^2}$$ Solve this equation to find the relation between \(r_1\) and \(r_2\) as follows:$$\frac{1}{r_1^2} = \frac{2}{r_2^2}$$
03

3. Find the ratio of distances

To find the ratio of distances \(r_1\) and \(r_2\), take the square root of both sides of the equation, giving:$$\frac{1}{r_1} = \sqrt{2}\frac{1}{r_2}$$ Then, find the ratio: $$\frac{r_1}{r_2} = \frac{1}{\sqrt{2}}$$
04

4. Analyze the ratio

The above equation shows that the point where the electric field is zero is located at a distance \(r_1\) from charge \(q\) and a distance \(r_2\) from charge \(2q\), such that the ratio of distances is \(1/\sqrt{2}\). This implies that the zero-field point lies closer to charge \(2q\) than to charge \(q\). Therefore, the answer is (D) Yes, between the two charges but closer to \(2q\).

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Most popular questions from this chapter

Two positive charges of magnitude \(q\) and \(2 q\) are fixed in place along the \(x\)-axis. Is there any place along the \(x\)-axis where the total field could be zero? (A) Yes, somewhere to the left of the charge \(q\) (B) Yes, somewhere to the right of the charge \(2 q\) (C) Yes, between the two charges but closer to \(q\) (D) Yes, between the two charges but closer to \(2 q\) (E) No, the field can never be zero

A test charge \(+q\) and a test charge \(-q\) are released midway between the two plates. Let the voltage of the top plate be \(V\) and the voltage of the bottom plate be 0 . The distance between the two plates is \(d\). Which of the following statements about the test charges are true? i. Both charges have gained a kinetic energy of \(|q| E d / 2\) when they hit the plates. ii. Charge \(+q\) is initially at a positive voltage and charge \(-q\) is initially at a negative voltage. iii. Charge \(+q\) initially has a positive potential energy and charge \(-q\) initially has a negative potential energy. iv. Both charges lose potential energy. (A) i only (B) i and ii (C) i and iii (D) iii and iv (E) i, iii and iv

Two identical point charges \(q_1\) and \(q_2\) are at a distance \(r\) apart. If the size of \(q_1\) is doubled and the distance between them tripled, the strength of the electrical force between them (A) goes up by a factor of 3 . (B) goes down by a factor of 3 . (C) goes down by a factor of 9 . (D) goes down by a factor of \(2 / 3\). (E) goes down by a factor of \(2 / 9\)

An electron is released from rest just above a very large, negatively charged sheet, which carries surface charge density \(\sigma=5 \mathrm{nC} / \mathrm{m}^2\). When the electron is \(0.5 \mathrm{~m}\) above the sheet its speed is most nearly (A) \(4 \times 10^8 \mathrm{~m} / \mathrm{s}\) (B) \(3 \times 10^8 \mathrm{~m} / \mathrm{s}\) (C) \(5 \times 10^7 \mathrm{~m} / \mathrm{s}\) (D) \(1 \times 10^7 \mathrm{~m} / \mathrm{s}\) (E) \(300 \mathrm{~m} / \mathrm{s}\)

A rectangular loop is situated in a region with a uniform magnetic field of \(0.1 \mathrm{~T}\) pointing into the page, as shown below. The length of the loop is \(20 \mathrm{~cm}\) and the width is \(10 \mathrm{~cm}\). a) What is the magnetic flux through the loop? b) If the value of \(B\) is increased from \(0.1 \mathrm{~T}\) to \(0.5 \mathrm{~T}\) in \(0.3 \mathrm{~s}\), what will be the EMF induced into the loop? What will be the direction of the induced current? Justify your answer. c) If the resistance of the loop is \(R=1 \Omega\), what is the value of the current? d) If the \(B\)-field is tilted at an angle \(\theta=15^{\circ}\) from the normal to the page, what is the flux through the loop? e) Give two ways that one could change the induced EMF in the loop, other than changing \(B\). f) Draw a rotation axis from left to right in the plane of the page that passes through the loop's center. The loop is rotated about this axis at a constant angular velocity. Write an expression for the flux in terms of the angular velocity \(\omega\) and the time \(t\).
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