3.10: The Electric Potential (Answers)

Conceptual Questions

1. No. We can only define potential energies for conservative fields.

3. No, though certain orderings may be simpler to compute.

5. The electric field strength is zero because electric potential differences are directly related to the field strength. If the potential difference is zero, then the field strength must also be zero.

7. Potential difference is more descriptive because it indicates that it is the difference between the electric potential of two points.

9. They are very similar, but potential difference is a feature of the system; when a charge is introduced to the system, it will have a potential energy which may be calculated by multiplying the magnitude of the charge by the potential difference.

11. An electron-volt is a volt multiplied by the charge of an electron. Volts measure potential difference, electron-volts are a unit of energy.

Problems

29. a. \(\displaystyle U=3.4J\);

b. \(\displaystyle \frac{1}{2}mv^2=kQ_1Q_2(\frac{1}{r_i}−\frac{1}{r_f})→v=750m/s\)

31. \(\displaystyle U=4.36×10^{−18}J\)

33. \(\displaystyle \frac{1}{2}m_ev^2_e=qV,\frac{1}{2}m_Hv^2_H=qV,\)so that \(\displaystyle \frac{m_ev^2_e}{m_Hv^2_H}=1\) or \(\displaystyle \frac{v_e}{v_H}=42.8\).

35. \(\displaystyle 1V=1J/C;1J=1N⋅m→1V/m=1N/C\)

37. a. \(\displaystyle V_{AB}=3.00kV\);

b. \(\displaystyle V_{AB}=7.50kV\)

39. a. \(\displaystyle V_{AB}=Ed→E=5.63kV/m\);

b. \(\displaystyle V_{AB}=563V\)

41. a. \(\displaystyle ΔK=qΔV\) and \(\displaystyle V_{AB}=Ed\),so that \(\displaystyle ΔK=800keV\);

b. \(\displaystyle d=25.0km\)

43. One possibility is to stay at constant radius and go along the arc from \(\displaystyle P_1\) to \(\displaystyle P_2\), which will have zero potential due to the path being perpendicular to the electric field. Then integrate from a to b: \(\displaystyle V_{ab}=αln(\frac{b}{a})\)

45. \(\displaystyle V=144V\)

47. \(\displaystyle V=\frac{kQ}{r}→Q=8.33×10^{−7}C\); The charge is positive because the potential is positive.

49. a. \(\displaystyle V=45.0MV\);

b. \(\displaystyle V=\frac{kQ}{r}→r=45.0m\);

c. \(\displaystyle ΔU=132MeV\)

51. \(\displaystyle V=kQ/r\); a. Relative to origin, find the potential at each point and then calculate the difference. \(\displaystyle ΔV=135×10^3V\);

b. To double the potential difference, move the point from 20 cm to infinity; the potential at 20 cm is halfway between zero and that at 10 cm.

Additional Problems

89. \(\displaystyle E_{tot}=4.67×10^7J\) \(\displaystyle E_{tot}=qV→q=\frac{E_{tot}}{V}=3.89×10^6C\)

91. \(\displaystyle V_P=k\frac{q_{tot}}{\sqrt{z^2+R^2}}→q_{tot}=−3.5×10^{−11}C\)

93. \(\displaystyle V_P=−2.2GV\)

95. Recall from the previous chapter that the electric field \(\displaystyle E_P=\frac{σ_0}{2ε_0}\) is uniform throughout space, and that for uniform fields we have \(\displaystyle E=−\frac{ΔV}{Δz}\) for the relation. Thus, we get \(\displaystyle \frac{σ}{2ε_0}=\frac{ΔV}{Δz}→Δz=0.22m\) for the distance between 25-V equipotentials.

97. a. Take the result from Example 7.13, divide both the numerator and the denominator by x, take the limit of that, and then apply a Taylor expansion to the resulting log to get: \(\displaystyle V_P≈kλ\frac{L}{x}\);

b. which is the result we expect, because at great distances, this should look like a point charge of \(\displaystyle q=λL\)

99. a. \(\displaystyle V=9.0×10^3V\);

b. \(\displaystyle −9.0×10^3V(\frac{1.25cm}{2.0cm})=−5.7×10^3V\)

101. a. \(\displaystyle E=\frac{KQ}{r^2}→Q=−6.76×10^5C\);

b. \(\displaystyle F=ma=qE→a=\frac{qE}{m}=2.63×10^{13}m/s^2(upwards)\);

c. \(\displaystyle F=−mg=qE→m=\frac{−qE}{g}=2.45×10^{−18}kg\)

103. If the electric field is zero ¼ from the way of \(\displaystyle q_1\) and \(\displaystyle q_2\), then we know from \(\displaystyle E=k\frac{Q}{r^2}\) that \(\displaystyle |E_1|=|E_2|→\frac{Kq_1}{x^2}=\frac{Kq^2}{(3x)^2}\) so that \(\displaystyle \frac{q_2}{q_1}=\frac{(3x)^2}{x^2}=9\); the charge \(\displaystyle q_2\) is 9 times larger than \(\displaystyle q_1\).

105. a. The field is in the direction of the electron’s initial velocity.

b. \(\displaystyle v^2=v^2_0+2ax→x=−\frac{v^2_0}{2a}(v=0)\). Also, \(\displaystyle F=ma=qE→a=\frac{qE}{m}\),\(\displaystyle x=3.56×10^{−4}m\);

c. \(\displaystyle v_2=v_0+at→t=−\frac{v_0m}{qE}(v=0), ∴t=1.42×10^{−10}s\);

d. \(\displaystyle v=−(\frac{2qEx}{m})^{1/2}−5.00×10^6m/s\) (opposite its initial velocity)

Challenge Problems

107. Answers will vary. This appears to be proprietary information, and ridiculously difficult to find. Speeds will be 20 m/s or less, and there are claims of \(\displaystyle ~10^{−7}\) grams for the mass of a drop.

109. Apply \(\displaystyle \vec{E}=−\vec{∇}V\) with \(\displaystyle \vec{∇}=\hat{r}\frac{∂}{∂r}+\hat{θ}\frac{1}{r}\frac{∂}{∂θ}+\hat{φ}\frac{1}{rsinθ}\frac{∂}{∂φ}\) to the potential calculated earlier, \(\displaystyle V_P=k\frac{\vec{p}⋅\hat{r}}{r^2}\) with \(\displaystyle \vec{p}=q\vec{d}\), and assume that the axis of the dipole is aligned with the z-axis of the coordinate system. Thus, the potential is \(\displaystyle V_P=k\frac{q\vec{d}⋅\hat{r}}{r^2}=k\frac{qdcosθ}{r^2}\).

\(\displaystyle \vec{E}=2kqd(\frac{cosθ}{r^3})\hat{r}+kqd(\frac{sinθ}{r^3})\hat{θ}\)

Contributors and Attributions

Samuel J. Ling (Truman State University), Jeff Sanny (Loyola Marymount University), and Bill Moebs with many contributing authors. This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0).