Q: An alpha particle ($$ \alpha^{2+} ) is $$accelerated from rest through a potential difference of $$100\ \mathrm{V}. $$What is its maximum speed? (Given: $$q_{\alpha} = 3.2 \times 10$$^{-19}$$\ \mathrm{C}$$, $$m_{\alpha} = 6.65 \times 10$$^{-27}$$\ \mathrm{kg})$$

$$v = \sqrt{\frac{2qV}{m}} = \sqrt{\frac{2 \times 3.2 \times 10$$^{-19}$$ \times 100}{6.65 \times 10$$^{-27}$$}} = 9.8 \times 10$$^{4}$$\ \mathrm{m/s}$$

Q: If an alpha particle moves from point A (10 V) to point B (30 V) in a uniform electric field, what is the change in its electric potential energy? ($$q_{\alpha} = 3.2 \times 10$$^{-19}$$\ \mathrm{C})$$

$$\Delta U = q \Delta V = 3.2 \times 10$$^{-19}$$ \times (30 - 10) = 6.4 \times 10$$^{-18}$$\ \mathrm{J}$$

Q: A $$7.0\ \mu\mathrm{C}$$ charge (mass $$2\ \mathrm{mg}) is $$released from rest at point B (30 V) and moves to point A (10 V). What is its final speed?

$$v = \sqrt{\frac{2q\Delta V}{m}} = \sqrt{\frac{2 \times 7.0 \times 10$$^{-6}$$ \times (30 - 10)}{2 \times 10$$^{-6}$$}} = 11.8\ \mathrm{m/s}$$

Q: A $$\mathrm{Be}^{2+}$$ ion is accelerated through a potential difference of $$0.150\ \mathrm{MV}$$ and reaches a speed of $$2.53 \times 10^6$\ \mathrm{m/s}. $$What is its mass? ($$q = 3.2 \times 10$$^{-19}$$\ \mathrm{C})$$

$$m = \frac{2qV}{v^2$} = \frac{2 \times 3.2 \times 10$$^{-19}$$ \times 1.5 \times 10$$^{5}$$}{(2.53 \times 10^6$)^2$$} = 1.5 \times $$10^{-26}$$\ \mathrm{kg}$

Q: A negatively charged particle is suspended between two plates $$20\ \mathrm{cm}$$ apart with a voltage of $$3.1 \times 10^2$\ \mathrm{V}$$ across them. If the electric force is $$8.4 \times 10$$^{-7}$$\ \mathrm{N}$$, what is the magnitude of the charge?

$$E = \frac{V}{d} = \frac{3.1 \times 10^2$}{0.2} = 1.55 \times 10^3$\ \mathrm{V/m}$$; $$q = \frac{F}{E} = \frac{8.4 \times 10$$^{-7}$$}{1.55 \times 10^3$} = 5.4 \times 10$$^{-10}$$\ \mathrm{C}$$

Q: A $$2.0\ \mu\mathrm{C}$$ charge moves from a point at $$50\ \mathrm{V} to a $$point at $$20\ \mathrm{V}. $$What is the change in electric potential energy?

$$\Delta U = q\Delta V = 2.0 \times 10$$^{-6}$$ \times (20 - 50) = -6.0 \times 10$$^{-5}$$\ \mathrm{J}$$

Q: A $$1.0\ \mathrm{mg}$$ particle with charge $$5.0 \times 10$$^{-6}$$\ \mathrm{C} is $$released from rest at $$40\ \mathrm{V}$$ and moves to $$0\ \mathrm{V}. $$What is its final speed?

$$v = \sqrt{\frac{2q\Delta V}{m}} = \sqrt{\frac{2 \times 5.0 \times 10$$^{-6}$$ \times 40}{1.0 \times 10$$^{-6}$$}} = 20\ \mathrm{m/s}$$

Q: A $$-2.0 \times 10$$^{-6}$$\ \mathrm{C}$$ charge is placed in a uniform electric field of $$1.0 \times 10$$^{4}$$\ \mathrm{N/C}. $$What is the force on the charge?

$$F = qE = -2.0 \times 10$$^{-6}$$ \times 1.0 \times 10$$^{4}$$ = -0.02\ \mathrm{N}$$

Question 1

An alpha particle ($$ \alpha^{2+} ) is $$accelerated from rest through a potential difference of $$100\ \mathrm{V}. $$What is its maximum speed? (Given: $$q_{\alpha} = 3.2 	imes 10$$^{-19}$$\ \mathrm{C}$$, $$m_{\alpha} = 6.65 	imes 10$$^{-27}$$\ \mathrm{kg})$$

Accepted Answer

$$v = \sqrt{\frac{2qV}{m}} = \sqrt{\frac{2 	imes 3.2 	imes 10$$^{-19}$$ 	imes 100}{6.65 	imes 10$$^{-27}$$}} = 9.8 	imes 10$$^{4}$$\ \mathrm{m/s}$$

Question 2

If an alpha particle moves from point A (10 V) to point B (30 V) in a uniform electric field, what is the change in its electric potential energy? ($$q_{\alpha} = 3.2 	imes 10$$^{-19}$$\ \mathrm{C})$$

Accepted Answer

$$\Delta U = q \Delta V = 3.2 	imes 10$$^{-19}$$ 	imes (30 - 10) = 6.4 	imes 10$$^{-18}$$\ \mathrm{J}$$

Question 3

A $$7.0\ \mu\mathrm{C}$$ charge (mass $$2\ \mathrm{mg}) is $$released from rest at point B (30 V) and moves to point A (10 V). What is its final speed?

Accepted Answer

$$v = \sqrt{\frac{2q\Delta V}{m}} = \sqrt{\frac{2 	imes 7.0 	imes 10$$^{-6}$$ 	imes (30 - 10)}{2 	imes 10$$^{-6}$$}} = 11.8\ \mathrm{m/s}$$

Question 4

A $$\mathrm{Be}^{2+}$$ ion is accelerated through a potential difference of $$0.150\ \mathrm{MV}$$ and reaches a speed of $$2.53 	imes 10^6$\ \mathrm{m/s}. $$What is its mass? ($$q = 3.2 	imes 10$$^{-19}$$\ \mathrm{C})$$

Accepted Answer

$$m = \frac{2qV}{v^2$} = \frac{2 	imes 3.2 	imes 10$$^{-19}$$ 	imes 1.5 	imes 10$$^{5}$$}{(2.53 	imes 10^6$)^2$$} = 1.5 	imes $$10^{-26}$$\ \mathrm{kg}$

Question 5

A negatively charged particle is suspended between two plates $$20\ \mathrm{cm}$$ apart with a voltage of $$3.1 	imes 10^2$\ \mathrm{V}$$ across them. If the electric force is $$8.4 	imes 10$$^{-7}$$\ \mathrm{N}$$, what is the magnitude of the charge?

Accepted Answer

$$E = \frac{V}{d} = \frac{3.1 	imes 10^2$}{0.2} = 1.55 	imes 10^3$\ \mathrm{V/m}$$; $$q = \frac{F}{E} = \frac{8.4 	imes 10$$^{-7}$$}{1.55 	imes 10^3$} = 5.4 	imes 10$$^{-10}$$\ \mathrm{C}$$

Question 6

Which of the following best describes the relationship between electric potential difference and kinetic energy for a charged particle accelerated from rest?

Accepted Answer

The kinetic energy gained is equal to the charge multiplied by the potential difference: $$KE = qV$$

Question 7

A $$2.0\ \mu\mathrm{C}$$ charge moves from a point at $$50\ \mathrm{V} to a $$point at $$20\ \mathrm{V}. $$What is the change in electric potential energy?

Accepted Answer

$$\Delta U = q\Delta V = 2.0 	imes 10$$^{-6}$$ 	imes (20 - 50) = -6.0 	imes 10$$^{-5}$$\ \mathrm{J}$$

Question 8

If a $$-3.0\ \mu\mathrm{C}$$ charge is suspended in equilibrium between two plates $$0.10\ \mathrm{m}$$ apart with a voltage of $$200\ \mathrm{V}$$, what is the magnitude of the electric force acting on it?

Accepted Answer

$$E = \frac{V}{d} = \frac{200}{0.10} = 2000\ \mathrm{V/m}$$; $$F = |q|E = 3.0 	imes 10$$^{-6}$$ 	imes 2000 = 6.0 	imes 10$$^{-3}$$\ \mathrm{N}$$

Question 9

A $$1.0\ \mathrm{mg}$$ particle with charge $$5.0 	imes 10$$^{-6}$$\ \mathrm{C} is $$released from rest at $$40\ \mathrm{V}$$ and moves to $$0\ \mathrm{V}. $$What is its final speed?

Accepted Answer

$$v = \sqrt{\frac{2q\Delta V}{m}} = \sqrt{\frac{2 	imes 5.0 	imes 10$$^{-6}$$ 	imes 40}{1.0 	imes 10$$^{-6}$$}} = 20\ \mathrm{m/s}$$

Question 10

A $$-2.0 	imes 10$$^{-6}$$\ \mathrm{C}$$ charge is placed in a uniform electric field of $$1.0 	imes 10$$^{4}$$\ \mathrm{N/C}. $$What is the force on the charge?

Accepted Answer

$$F = qE = -2.0 	imes 10$$^{-6}$$ 	imes 1.0 	imes 10$$^{4}$$ = -0.02\ \mathrm{N}$$

Question 11

Which of the following statements about equipotential lines is correct?

Accepted Answer

No work is required to move a charge along an equipotential line.

Question 12

A $$1.0 	imes 10$$^{-6}$$\ \mathrm{C}$$ charge is moved from a point at $$20\ \mathrm{V} to a $$point at $$50\ \mathrm{V}. $$How much work is done by the electric field?

Accepted Answer

$$W = -q\Delta V = -1.0 	imes 10$$^{-6}$$ 	imes (50 - 20) = -3.0 	imes 10$$^{-5}$$\ \mathrm{J}$$

Electric Potential, Energy, and Motion of Charged Particles in Electric Fields

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