November 15, 2019

Solution Set 8

Physics 208-02 – Introductory Physics II
Solution Set 8
Nazareth College
Department of Chemistry & Biochemistry
Robert F Szalapski, PhD – Adjunct Lecturer
www.CallMeDrRob.com
Spring 2012

§ Chapter 17, Problem 4

An electron acquires 7.45\times 10^{{-16}}\mathrm{J} of kinetic energy when it is accelerated by an electric field from plate A to plate B. What is the potential difference between the plates, and which is at a higher potential?

An electron is negatively charged, hence it accelerates in the direction opposite to the electric field. The electric field then points from B towards A, so B would be at the higher potential. Let’s choose our zero of potential energy as plate A, and then B is at some unknown potential V_{0}. (Only potential differences actually matter.) Assume that the electron was initially at rest.

\displaystyle\left(KE\right)_{\mathrm{initial}}+\left(PE\right)_{\mathrm{initial}} \displaystyle= \displaystyle\left(KE\right)_{\mathrm{final}}+\left(PE\right)_{\mathrm{final}}
\displaystyle 0+(-e)(0) \displaystyle= \displaystyle\left(KE\right)_{\mathrm{final}}+(-e)V_{0}
\displaystyle V_{0} \displaystyle= \displaystyle\frac{\left(KE\right)_{\mathrm{final}}}{e}
\displaystyle= \displaystyle\frac{7.45\times 10^{{-16}}\mathrm{J}}{1.602\times 10^{{-19}}\mathrm{C}}
\displaystyle= \displaystyle 4.65\;\mathrm{kV}

§ Chapter 17, Problem 6

An electric field of 640\;\mathrm{V}/\mathrm{m} is required between two parallel plates 11.0\;\mathrm{mm} apart. How large a voltage should be applied?

\displaystyle E \displaystyle= \displaystyle\frac{\Delta V}{d}
\displaystyle\Delta V \displaystyle= \displaystyle Ed
\displaystyle= \displaystyle\left(640\frac{\mathrm{V}}{\mathrm{m}}\right)\left(11.0\times 10^{{-3}}\mathrm{m}\right)
\displaystyle= \displaystyle 7.0\;\mathrm{V}

§ Chapter 17, Problem 8

What potential difference is needed to give a helium nucleus (Q=2e) 65.0\;\mathrm{keV} of kinetic energy?

This is like Problem 4, except the nucleus will accelerate from the higher positive potential in the direction of the electric field lines because it is positively charged. Assume it is initially at rest.

\displaystyle\left(KE\right)_{\mathrm{initial}}+\left(PE\right)_{\mathrm{initial}} \displaystyle= \displaystyle\left(KE\right)_{\mathrm{final}}+\left(PE\right)_{\mathrm{final}}
\displaystyle 0+(2e)V_{0} \displaystyle= \displaystyle\left(KE\right)_{\mathrm{final}}+(2e)(0)
\displaystyle V_{0} \displaystyle= \displaystyle\frac{\left(KE\right)_{\mathrm{final}}}{2e}
\displaystyle= \displaystyle\frac{65.0\times 10^{3}e\mathrm{V}}{2e}
\displaystyle= \displaystyle 32.5kV

Notice how we took advantage of the definition of an electron volt!

§ Chapter 17, Problem 11

What is the speed of an electron with energy:

§ (a)

750\;\mathrm{eV}?

\displaystyle\frac{1}{2}m_{e}v^{2} \displaystyle= \displaystyle 750\;\mathrm{eV}
\displaystyle= \displaystyle e(750\;\mathrm{V})
\displaystyle v \displaystyle= \displaystyle\sqrt{\frac{2e}{m_{e}}(750\;\mathrm{V})}
\displaystyle= \displaystyle\sqrt{2\frac{1.602\times 10^{{-19}}\mathrm{C}}{9.11\times 10^{{-31}}\mathrm{kg}}(750\;\mathrm{V})}
\displaystyle= \displaystyle 1.6\times 10^{7}\frac{\mathrm{m}}{\mathrm{s}}

§ (b)

3.2\;\mathrm{keV}?

\displaystyle\frac{1}{2}m_{e}v^{2} \displaystyle= \displaystyle 3.2\times 10^{3}\mathrm{eV}
\displaystyle= \displaystyle e\left(3.2\times 10^{3}\mathrm{eV}\right)
\displaystyle v \displaystyle= \displaystyle\sqrt{\frac{2e}{m_{e}}\left(3.2\times 10^{3}\mathrm{eV}\right)}
\displaystyle= \displaystyle\sqrt{2\frac{1.602\times 10^{{-19}}\mathrm{C}}{9.11\times 10^{{-31}}\mathrm{kg}}\left(3.2\times 10^{3}\mathrm{eV}\right)}
\displaystyle= \displaystyle 3.4\times 10^{7}\frac{\mathrm{m}}{\mathrm{s}}

§ Chapter 17, Problem 12

What is the speed of a proton whose kinetic energy is 3.2\;\mathrm{keV}?.

\displaystyle\frac{1}{2}m_{p}v^{2} \displaystyle= \displaystyle 3.2\times 10^{3}\mathrm{eV}
\displaystyle= \displaystyle e\left(3.2\times 10^{3}\mathrm{eV}\right)
\displaystyle v \displaystyle= \displaystyle\sqrt{\frac{2e}{m_{p}}\left(3.2\times 10^{3}\mathrm{eV}\right)}
\displaystyle= \displaystyle\sqrt{2\frac{1.602\times 10^{{-19}}\mathrm{C}}{1.67\times 10^{{-27}}\mathrm{kg}}\left(3.2\times 10^{3}\mathrm{eV}\right)}
\displaystyle= \displaystyle 7.8\times 10^{5}\frac{\mathrm{m}}{\mathrm{s}}

§ Chapter 17, Problem 14

What is the electric potential 15.0\;\mathrm{cm} from a 4.00\;\mu\mathrm{C} charge?

\displaystyle V \displaystyle= \displaystyle k\frac{q}{r}
\displaystyle= \displaystyle\left(9.0\times 10^{9}\frac{\mathrm{N}\,\mathrm{m}^{2}}{\mathrm{C}^{2}}\right)\frac{4.00\times 10^{{-6}}\mathrm{C}}{0.150\;\mathrm{m}}
\displaystyle= \displaystyle 2.40\times 10^{5}\;\mathrm{V}

§ Chapter 17, Problem 15

A point charge Q creates an electric potential of +125\;\mathrm{V} at a distance of 15\;\mathrm{cm}. What is Q?

\displaystyle V \displaystyle= \displaystyle k\frac{Q}{r}
\displaystyle Q \displaystyle= \displaystyle V\frac{r}{k}
\displaystyle= \displaystyle 125\;\mathrm{V}\frac{0.15\;\mathrm{m}}{9.0\times 10^{9}\frac{\mathrm{N}\,\mathrm{m}^{2}}{\mathrm{C}^{2}}}
\displaystyle= \displaystyle 2.1\;\mathrm{nC}

§ Chapter 17, Problem 16

A +35\;\mu\mathrm{C} charge is placed 32\;\mathrm{cm} from an identical +35\;\mu\mathrm{C} charge. How much work would be required to move a +0.50\;\mu\mathrm{C} charge 12\;\mathrm{cm} closer to either of the charges?

See the figure. This problem would seem difficult without a sketch. To proceed, calculate the potential due to the two charges at the point where the test charge will be placed. This is much easier than an electric-field problem because we are dealing with scalars, not vectors.

\displaystyle V_{\mathrm{before}} \displaystyle= \displaystyle k\frac{q_{1}}{r_{1}}+k\frac{q_{2}}{r_{2}}
\displaystyle= \displaystyle\left(9.0\times 10^{9}\frac{\mathrm{N}\,\mathrm{m}^{2}}{\mathrm{C}^{2}}\right)\frac{35\times 10^{{-6}}\mathrm{C}}{0.16\;\mathrm{m}}+\left(9.0\times 10^{9}\frac{\mathrm{N}\,\mathrm{m}^{2}}{\mathrm{C}^{2}}\right)\frac{35\times 10^{{-6}}\mathrm{C}}{0.16\;\mathrm{m}}
\displaystyle= \displaystyle 3.\underline{9}4\times 10^{6}\,\mathrm{V}
\displaystyle V_{\mathrm{after}} \displaystyle= \displaystyle k\frac{q_{1}}{r_{1}}+k\frac{q_{2}}{r_{2}}
\displaystyle= \displaystyle\left(9.0\times 10^{9}\frac{\mathrm{N}\,\mathrm{m}^{2}}{\mathrm{C}^{2}}\right)\frac{35\times 10^{{-6}}\mathrm{C}}{0.04\;\mathrm{m}}+\left(9.0\times 10^{9}\frac{\mathrm{N}\,\mathrm{m}^{2}}{\mathrm{C}^{2}}\right)\frac{35\times 10^{{-6}}\mathrm{C}}{0.28\;\mathrm{m}}
\displaystyle= \displaystyle\underline{9}.0\times 10^{6}\,\mathrm{V}

We have moved the test charge to a higher potential. Next let’s compute the potential energy at each point.

\displaystyle\mathrm{PE} \displaystyle= \displaystyle qV
\displaystyle\mathrm{PE}_{\mathrm{before}} \displaystyle= \displaystyle\left(+0.50\times 10^{{-6}}\,\mathrm{C}\right)\left(3.\underline{9}4\times 10^{6}\,\mathrm{V}\right)
\displaystyle= \displaystyle 2.0\;\mathrm{J}
\displaystyle\mathrm{PE}_{\mathrm{after}} \displaystyle= \displaystyle\left(+0.50\times 10^{{-6}}\,\mathrm{C}\right)\left(\underline{9}.0\times 10^{6}\,\mathrm{V}\right)
\displaystyle= \displaystyle\underline{4}.5\;\mathrm{J}

We have moved the test charge to a position of greater potential energy, and hence we will need to do positive work. (The force is in the direction of the displacement.)

\displaystyle W \displaystyle= \displaystyle\underline{4}.5\;\mathrm{J}-2.0\;\mathrm{J}
\displaystyle= \displaystyle\underline{2}.5\;\mathrm{J}

§ Chapter 17, Problem 18

§ (a)

What is the electric potential a distance of 2.5\times 10^{{-15}}\,\mathrm{m} away from a proton?

\displaystyle V \displaystyle= \displaystyle k\frac{q}{r}
\displaystyle= \displaystyle\left(9.0\times 10^{9}\frac{\mathrm{N}\,\mathrm{m}^{2}}{\mathrm{C}^{2}}\right)\frac{1.602\times 10^{{-19}}\,\mathrm{C}}{2.5\times 10^{{-15}}\mathrm{m}}
\displaystyle= \displaystyle 5.8\times 10^{5}\,\mathrm{V}

§ (b)

What is the electric potential energy of a system of two protons a distance of 2.5\times 10^{{-15}}\,\mathrm{m} apart, as occurs in a typical nucleus?

\displaystyle\mathrm{PE} \displaystyle= \displaystyle qV
\displaystyle= \displaystyle\left(1.602\times 10^{{-19}}\,\mathrm{C}\right)\left(5.8\times 10^{5}\,\mathrm{V}\right)
\displaystyle= \displaystyle 9.2\times 10^{{-14}}\,\mathrm{J}

§ Chapter 17, Problem 20

An electron starts from rest 32.5\;\mathrm{cm} from a fixed point charge with Q=-0.125\;\mu\mathrm{C}. How fast will the electron be when it is very far away?

Initially there is no kinetic energy. Compute the potential energy. Initially compute the potential due to the charge Q at the position of the electron.

\displaystyle V \displaystyle= \displaystyle k\frac{q}{r}
\displaystyle= \displaystyle\left(9.0\times 10^{9}\frac{\mathrm{N}\,\mathrm{m}^{2}}{\mathrm{C}^{2}}\right)\frac{\left(-0.125\times 10^{{-6}}\,\mathrm{C}\right)}{0.325\;\mathrm{m}}
\displaystyle= \displaystyle-3.46\times 10^{3}\,\mathrm{V}

Now compute the potential energy.

\displaystyle\mathrm{PE} \displaystyle= \displaystyle qV
\displaystyle= \displaystyle-eV
\displaystyle= \displaystyle\left(-1.602\times 10^{{-19}}\,\mathrm{C}\right)\left(-3.46\times 10^{3}\,\mathrm{V}\right)
\displaystyle= \displaystyle 5.55\times 10^{{-16}}\,\mathrm{J}

When the electron is very far from Q the potential energy will be zero, and all of the energy will be kinetic energy.

\displaystyle\frac{1}{2}m_{e}v^{2} \displaystyle= \displaystyle E
\displaystyle v \displaystyle= \displaystyle\sqrt{\frac{2E}{m_{e}}}
\displaystyle= \displaystyle\frac{2\left(5.55\times 10^{{-16}}\,\mathrm{J}\right)}{9.11\times 10^{{-31}}\,\mathrm{kg}}
\displaystyle= \displaystyle 3.49\times 10^{7}\frac{\mathrm{m}}{\mathrm{s}}

§ Chapter 17, Problem 21

Two identical +9.5\;\mu\mathrm{C} point charges are initially 3.5\;\mathrm{cm} from each other. If they are released at the same instant from rest, how fast will each be moving when they are very far away from each other? Assume that they have identical masses of 1.0\;\mathrm{mg}.

Compute the potential due to one at the position of the other.

\displaystyle V \displaystyle= \displaystyle k\frac{q}{r}
\displaystyle= \displaystyle\left(9.0\times 10^{9}\frac{\mathrm{N}\,\mathrm{m}^{2}}{\mathrm{C}^{2}}\right)\frac{\left(+9.5\times 10^{{-6}}\,\mathrm{C}\right)}{0.035\;\mathrm{m}}
\displaystyle= \displaystyle 2.4\times 10^{6}\,\mathrm{V}

Now compute the potential energy.

\displaystyle\mathrm{PE} \displaystyle= \displaystyle qV
\displaystyle= \displaystyle\left(+9.5\times 10^{{-6}}\,\mathrm{C}\right)\left(2.4\times 10^{6}\,\mathrm{V}\right)
\displaystyle= \displaystyle 23\;\mathrm{J}

This energy will be distributed equally between the kinetic energies of the two particles when they have moved apart in a symmetric fashion to a very large distance where the potential energy is effectively zero.

\displaystyle\frac{1}{2}\left(1.0\times 10^{{-6}}\,\mathrm{kg}\right)v^{2} \displaystyle= \displaystyle\frac{1}{2}\left(23\;\mathrm{J}\right)
\displaystyle= \displaystyle 4.8\times 10^{3}\frac{\mathrm{m}}{\mathrm{s}}

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