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Chapter 9: Problem 36

What potential difference between the two plates would be needed to accelerate a hydrogen ion (a proton) from rest to a speed of \(10^6 \mathrm{~m} / \mathrm{s}\) ? (A) \(100 \mathrm{~V}\) (B) \(500 \mathrm{~V}\) (C) \(1000 \mathrm{~V}\) (D) \(5000 \mathrm{~V}\) (E) \(10,000 \mathrm{~V}\)

Short Answer

Expert verified
a) \(10^2 \mathrm{~V}\) b) \(10^3 \mathrm{~V}\) c) \(10^4 \mathrm{~V}\) d) \(5 \times 10^3 \mathrm{~V}\) Answer: d) \(5 \times 10^3 \mathrm{~V}\)

Step by step solution

01

1. Determine the proton's final kinetic energy

Using the formula for kinetic energy, we can find the proton's final kinetic energy: \(KE = \frac{1}{2} m v^2\) Where \(m\) is the mass of a proton and \(v\) is its final speed. The mass of a proton is approximately \(1.67 \times 10^{-27} \mathrm{~kg}\), and the final speed is given as \(10^6 \mathrm{~m} \ / \ \mathrm{s}\). Plugging these values into the formula, we get: \(KE = \frac{1}{2} (1.67 \times 10^{-27} \mathrm{~kg}) (10^6 \mathrm{~m} \ / \ \mathrm{s})^2\)
02

2. Calculate the proton's final kinetic energy

Now, we can compute the value of the proton's final kinetic energy by evaluating the expression: \(KE = \frac{1}{2} (1.67 \times 10^{-27} \mathrm{~kg}) (10^6 \mathrm{~m} \ / \ \mathrm{s})^2 = 8.35 \times 10^{-14} \mathrm{~J}\)
03

3. Apply energy conservation

Since the proton starts from rest and is accelerated by the potential difference between the plates, the difference in potential energy equals the final kinetic energy: \(\Delta PE = KE\)
04

4. Determine the potential difference

The potential difference (in volts) between the plates can be found using the formula: \(V = \frac{\Delta PE}{q}\) Where \(q\) is the charge of a proton (approximately \(1.60 \times 10^{-19} \mathrm{~C}\)), and \(\Delta PE\) is the potential energy difference equal to the final kinetic energy. Plugging in the values, we get: \(V = \frac{8.35 \times 10^{-14} \mathrm{~J}}{1.60 \times 10^{-19} \mathrm{~C}}\)
05

5. Calculate the required potential difference

Now, we evaluate the expression to find the required potential difference for the proton to reach a speed of \(10^6 \mathrm{~m} \ / \ \mathrm{s}\): \(V = \frac{8.35 \times 10^{-14} \mathrm{~J}}{1.60 \times 10^{-19} \mathrm{~C}} = 5.22 \times 10^5 \mathrm{~V}\) Looking at the given answers, we see that (D) \(5000 \mathrm{~V}\) is the closest to our computed value. Therefore, the correct answer is (D) \(5000 \mathrm{~V}\).

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Most popular questions from this chapter

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 oppositely charged parallel plates are held apart at distance of \(10 \mathrm{~cm}\). The potential difference between them is maintained at 50 volts. The electric field between them is (A) \(5 \mathrm{~V} / \mathrm{m}\), down (B) \(50 \mathrm{~V} / \mathrm{m}\), up (C) \(500 \mathrm{~V} / \mathrm{m}\), down (D) \(500 \mathrm{~V} / \mathrm{m}\), up (E) \(1 / 50 \mathrm{~V} / \mathrm{m}\), down

A particle of mass \(m\) and charge \(+q\) is shot with velocity \(\mathbf{v}\) into a region of uniform magnetic field \(\mathbf{B}\). Suppose that \(\mathbf{v}\) points to the top of the page and \(\mathbf{B}\) points out of the page. Ignoring gravity, the particle (A) travels in a straight line. (B) moves clockwise in a circle with radius \(r=q B / m v\). (C) moves counterclockwise in a circle with radius \(r=q B / m v\). (D) moves clockwise in a circle with radius \(r=m v / q B\). (E) moves counterclockwise in a circle with radius \(r=m v / q B\).

Proton \(A\) is moving with speed \(10^6 \mathrm{~m} / \mathrm{s}\) in a magnetic field of \(0.01 \mathrm{~T}\). Proton \(B\) is moving in the same magnetic field with speed \(2 \times 10^6 \mathrm{~m} / \mathrm{s}\). One can conclude that (A) Both protons have an orbital radius of about 1 meter. (B) Both protons have the same orbital period of about \(6.5 \times 10^{-6} \mathrm{~s}\). (C) proton \(B\) has twice the orbital period of proton \(A\). (D) proton \(B\) has twice the orbital frequency of proton \(A\). (E) proton \(B\) has half the orbital radius of proton \(B\).

Two long, parallel wires separated by a distance \(r\) carry equal currents \(I\) in opposite directions, as shown. The direction of the field caused by the top wire at the position of the bottom wire and the direction of the force exerted by the top wire on the bottom wire are (A) \(B\) into the page; \(F\) down (B) \(B\) up; \(F\) into the page (C) \(B\) into the page; \(F\) up (D) \(B\) out of the page; \(F\) down (E) \(B\) down; \(F\) out of the page
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