16.7: Direct Calculation of Electrical Quantities from Charge Distributions (Answers)

Note: Answers are provided for only the odd-numbered questions.

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.

$\displaystyle dF_y=\frac{1}{4πε_0}\frac{aqλdx}{(x^2+a^2)^{3/2}}$,

$\displaystyle Fy=\frac{1}{2πε_0}\frac{qλ}{a}[\frac{l/2}{((l/2)^2+a^2)^{1/2}}]$