4.8: Potential and Field Relationships (Answers)

Conceptual Questions

17. No. It will be constant, but not necessarily zero.

19. no

21. No; it might not be at electrostatic equilibrium.

23. Yes. It depends on where the zero reference for potential is. (Though this might be unusual.)

25. So that lightning striking them goes into the ground instead of the television equipment.

Problems

59. a. increases; the constant (negative) electric field has this effect, the reference point only matters for magnitude; b. they are planes parallel to the sheet; c. 0.006 m

61. a. from the previous chapter, the electric field has magnitude \(\displaystyle \frac{σ}{ε_0}\) in the region between the plates and zero outside; defining the negatively charged plate to be at the origin and zero potential, with the positively charged plate located at +5mm in the z-direction, V=1.7×104VV=1.7×104Vso the potential is 0 for \(\displaystyle z<0,1.7×10^4V(\frac{z}{5mm})\) for \(\displaystyle 0≤z≤5mm,1.7×10^4V\) for \(\displaystyle z>5mm;\)

b. \(\displaystyle qV=\frac{1}{2}mv^2→v=7.7×10^7m/s\)

63. \(\displaystyle V=85V\)

65. In the region \(\displaystyle a≤r≤b, \vec{E}=\frac{kQ}{r^2}\hat{r}\), and E is zero elsewhere; hence, the potential difference is \(\displaystyle V=kQ(\frac{1}{a}−\frac{1}{b})\).

67. From previous results \(\displaystyle V_P−V_R=−2kλln\frac{s_P}{s_R}\)., note that b is a very convenient location to define the zero level of potential: \(\displaystyle ΔV=−2k\frac{Q}{L}ln\frac{a}{b}\).

69. a. \(\displaystyle F=5.58×10^{−11}N/C\);The electric field is towards the surface of Earth.

b. The coulomb force is much stronger than gravity.

71. We know from the Gauss’s law chapter that the electric field for an infinite line charge is \(\displaystyle \vec{E}_P=2kλ\frac{1}{s}\hat{s}\), and from earlier in this chapter that the potential of a wire-cylinder system of this sort is \(\displaystyle V_P=−2kλln\frac{s_P}{R}\) by integration. We are not given \(\displaystyle λ\), but we are given a fixed \(\displaystyle V_0\); thus, we know that \(\displaystyle V_0=−2kλln\frac{a}{R}\) and hence \(\displaystyle λ=−\frac{V_0}{2kln(\frac{a}{R})}\). We may substitute this back in to find a. \(\displaystyle \vec{E}_P=−\frac{V_0}{ln(\frac{a}{R})}\frac{1}{s}\hat{s}\);

b. \(\displaystyle V_P=V_0\frac{ln(\frac{sP}{R})}{ln(\frac{a}{R})}\);

c. \(\displaystyle 4.74×10^4N/C\)

73. a. \(\displaystyle U_1=7.68×10^{−18}J, U_2=5.76×10^{−18}J\);

b. \(\displaystyle U_1+U_2=−1.34×10^{−17}J\)

75. a. \(\displaystyle U=2.30×10^{−16}J;

b. \(\displaystyle \overline{K}=\frac{3}{2}kT→T=1.11×10^7\)

77. a. \(\displaystyle 1.9×10^6m/s\);

b. \(\displaystyle 4.2×10^6m/s\);

c. \(\displaystyle 5.9×10^6m/s\);

d. \(\displaystyle 7.3×10^6m/s\);

e. \(\displaystyle 8.4×10^6m/s\)

79. a. \(\displaystyle E=2.5×10^6V/m<3×10^6V/m\) No, the field strength is smaller than the breakdown strength for air.

b. \(\displaystyle d=1.7mm\)

81. \(\displaystyle K_f=qV_{AB}=qEd→E=8.00×10^5V/m\)

83. a. Energy=\(\displaystyle 2.00×10^9J\);

b. \(\displaystyle Q=m(cΔT+L_∇)\) \(\displaystyle m=766kg\);

c. The expansion of the steam upon boiling can literally blow the tree apart.

85. a. \(\displaystyle V=\frac{kQ}{r}→r=1.80km\);

b. A 1-C charge is a very large amount of charge; a sphere of 1.80 km is impractical.

87. The alpha particle approaches the gold nucleus until its original energy is converted to potential energy. \(\displaystyle 5.00MeV=8.00×10^{−13}J\), so \(\displaystyle E_0=\frac{qkQ}{r}→r=4.54×10^{−14}m\)

(Size of gold nucleus is about \(\displaystyle 7×10^{−15}m\)).

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).