16.7: Direct Calculation of Electrical Quantities from Charge Distributions (Answers)
- Page ID
- 100313
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Conceptual Questions
Calculating Electric Potential of Charge Distributions
13. The second has 1/4 the dipole moment of the first.
15. The region outside of the sphere will have a potential indistinguishable from a point charge; the interior of the sphere will have a different potential.
Problems
Electric Dipoles
105. \(\displaystyle E_x=0, E_y=\frac{1}{4πε_0}[\frac{2q}{(x^2+a^2})\frac{a}{\sqrt{(x^2+a^2)}}⇒x≫a⇒\frac{1}{2πε_0}\frac{qa}{x^3}\)
\(\displaystyle E_y=\frac{q}{4πε_0}[\frac{2ya+2ya}{(y−a)^2(y+a)^2}]⇒y≫a⇒\frac{1}{πε_0}\frac{qa}{y^3}\)
107. The net dipole moment of the molecule is the vector sum of the individual dipole moments between the two O-H. The separation O-H is 0.9578 angstroms:
\(\displaystyle \vec{p} =1.889×10^{−29}Cm\hat{i}\)
Calculating Electric Fields of Charge Distributions
83. \(\displaystyle dE=\frac{1}{4πε_0}\frac{λdx}{(x+a)^2},E=\frac{λ}{4πε_0}[\frac{1}{l+a}−\frac{1}{a}]\)
87. At \(\displaystyle P_1: \vec{E}(y)=\frac{1}{4πε_0}\frac{λL}{y\sqrt{y^2+\frac{L^2}{4}}}\hat{j}⇒\frac{1}{4πε_0}\frac{q}{\frac{a}{2}\sqrt{(\frac{a}{2})^2+\frac{L^2}{4}}}\hat{j}=\frac{1}{πε_0}\frac{q}{a\sqrt{a^2+L^2}}\hat{j}\)
At \(\displaystyle P_2\): Put the origin at the end of L.
\(\displaystyle dE=\frac{1}{4πε_0}\frac{λdx}{(x+a)^2},\vec{E} =−\frac{q}{4πε_0l}[\frac{1}{l+a}−\frac{1}{a}]\hat{i}\)
97. circular arc \(\displaystyle dE_x(−\hat{i})=\frac{1}{4πε_0}\frac{λds}{r^2}cosθ(−\hat{i}\),
\(\displaystyle \vec{E}_x=\frac{λ}{4πε_0r}(−\hat{i})\),
\(\displaystyle dEy(−\hat{i}ˆ)=\frac{1}{4πε_0}\frac{λds}{r^2}sinθ(−\hat{j})\),
\(\displaystyle \vec{E}_y=\frac{λ}{4πε_0r}(−\hat{j})\);
y-axis: \(\displaystyle \vec{E}_x=\frac{λ}{4πε_0r}(−\hat{i})\);
x-axis: \(\displaystyle \vec{E}_y=\frac{λ}{4πε_0r}(−\hat{j})\),
\(\displaystyle \vec{E}=\frac{λ}{2πε_0r}(−\hat{i})+\frac{λ}{2πε_0r}(−\hat{j})\)
Additional Problems
121. Electric field of wire at x: \(\displaystyle \vec{E}(x)=\frac{1}{4πε_0}\frac{2λ_y}{x}\hat{i}\),
\(\displaystyle dF=\frac{λ_yλ_x}{2πε_0}(lnb−lna)\)
123.
\(\displaystyle dEx=\frac{1}{4πε_0}\frac{λdx}{(x^2+a^2)}\frac{x}{\sqrt{x^2+a^2}}\),
\(\displaystyle \vec{E}_x=\frac{λ}{4πε_0}[\frac{1}{\sqrt{L^2+a^2}}−\frac{1}{a}]\hat{i}\),
\(\displaystyle dE_z=\frac{1}{4πε_0}\frac{λdx}{(x^2+a^2)}\frac{a}{\sqrt{x^2+a^2}}\),
\(\displaystyle \vec{E}_z=\frac{λ}{4πε_0a}\frac{L}{\sqrt{L^2+a^2}}\hat{k}\),
Substituting z for a, we have:
\(\displaystyle \vec{E}(z)=\frac{λ}{4πε_0}[\frac{1}{\sqrt{L^2+z^2}}−\frac{1}{z}]\hat{i}+\frac{λ}{4πε_0z}\frac{L}{\sqrt{L^2+z^2}}\hat{k}\)
125. There is a net force only in the y-direction. Let \(\displaystyle θ\) be the angle the vector from dx to q makes with the x-axis. The components along the x-axis cancel due to symmetry, leaving the y-component of the force.
\(\displaystyle dF_y=\frac{1}{4πε_0}\frac{aqλdx}{(x^2+a^2)^{3/2}}\),
\(\displaystyle Fy=\frac{1}{2πε_0}\frac{qλ}{a}[\frac{l/2}{((l/2)^2+a^2)^{1/2}}]\)