THE UNIVERSITY OF BRITISH COLUMBIA
 
Physics 122 Assignment # 3 SOLUTIONS:
 
ELECTRIC CHARGE
 
Wed. 16 Jan. 2002 - finish by Fri. 25 Jan.

1.

$\parbox{4.25in}{% {\bf TRIANGLE of CHARGES: } Derive an expression . . . . . . tre of the triangle. (Initially the charges are all infinitely far apart.) }$

$\parbox{4.5in}{% \raggedright\noindent{\bf ANSWER}:~ {\sf {% The wor . . . . . . r_{14}} + {q_2 q_4 \over r_{24}} + {q_3 q_4 \over r_{34}} \right] }$ , }}}$
 
where the labels ``1,2,3,4'' are shown on the Figure. (Note that no pair is counted twice.) Now, q₁ = q₄ = +q and q₂ = q₃ = -q; meanwhile r₁₂ = r₂₃ = r₁₃ = a and $r_{14} = r_{24} = r_{34} = a/\sqrt{3}$. So $\ds{ U_E = {q^2 \over 4\pi \epsilon_0 a} \left[ -1 +1 -1 + \sqrt{3} - \sqrt{3} - \sqrt{3} \right] }$ or Numerically, (Either form is acceptable.)

2.

LINES of CHARGE: An infinite line of charge with linear charge density $\lambda = 5 % \times 10^{-10}$ C/m lies along the x-axis (y=0, z=0 and $-\infty < x < +\infty$). A second infinite line of charge with exactly the opposite charge density lies along the z axis. What are the x, y and z components of the resultant electric field at the point (x,y,z) = (4,4,4) <tex2html_comment_mark> m?

ANSWER:  The text gives the result $\Vec{E} = \Hat{r} \lambda / 2 \pi \epsilon_0 r$ for the electric field due to a line of charge, where the $\Vec{r}$ is a vector from the line to the point at which $\Vec{E}$ is to be evaluated, perpendicular to the line. (By symmetry, there is no other possible direction for $\Vec{E}$ to point!) Thus the field $\Vec{E}_1$ due to the first line of charge points directly away from the x axis [at the specified point this will be in a direction $\Hat{r}_1 = (\jH + \kH)/\sqrt{2}$] and has a magnitude $E_1 = \lambda / 2 \pi \epsilon_0 r_1$ with $r_1 = \sqrt{4^2 + 4^2} = 4\sqrt{2}$ m, so $E_1 = {2 \times 5 \times 10^{-10} \over 2 \pi \epsilon_0 \times 4\sqrt{2} } = 1.589$ N/C. The second line of charge is the same distance away from our test point (r₂ = r₁) and the magnitude of the charge per unit length is the same $(\lambda)$ so the magnitude of the second contribution to the electric field is the same as that of the first (E₂ = E₁) but in this case the charge is opposite (negative) so $\Vec{E}_2$ points toward the other line: $\Vec{E}_2 = - \Hat{r}_2 E_2$ where $\Hat{r}_2 = (\iH + \jH)/\sqrt{2}$. Thus in total $\Vec{E} = \Vec{E}_1 + \Vec{E}_2 = [\jH + \kH - \iH - \jH] \times 1.589 / \sqrt{2}$ or

3.

$\parbox{3.25in}{% {\bf PLANE of CHARGE: } A large flat non-conducti . . . . . . re of the hole along its axis. The radius of the hole is $R = 0.4 % $ ~m. }$

 
ANSWER:  We can treat this as the superposition of a uniformly charged sheet (without a hole), which gives a uniform electric field $\ds{ E_1 = {\sigma \over 2 \epsilon_0} }$ normal to the surface, and a negatively charged disc at the position of the hole, which produces a field $\ds{ E_2 = - {\sigma \over 2 \epsilon_0} \left[ 1 - { z \over \sqrt{z^2 + R^2} } \right] }$ (parallel to the axis) at a point z away on the axis. Thus $\ds{ E = E_1 + E_2 = {\sigma \over 2 \epsilon_0} \left( 1 - \left[ 1 - { z \o . . . . . . 2 \times 0.8854 \times 10^{-11} } \times { 1.0 \over \sqrt{1.0^2 + 0.4^2} } }$ or directed along the positive z axis.

4.

EXTRA ELECTRONS: The Earth can be regarded as a spherical conductor from the point of view of free electrons. How many excess electrons could the Earth hold? (Assume that only electrostatic and gravitational forces are involved.)

ANSWER:  The N excess electrons repel each other and so ``try'' to get as far from one another as possible; they therefore accumulate uniformly over the surface of the Earth (radius R<sub>E</sub>). By the SHELL THEOREM (which works just as well for the electrostatic force as for the gravitational force, since both obey inverse square laws) the resultant electric field will be the same as if all the charge were concentrated at the Earth's centre. Thus the electrical force on any one electron will be $\ds{ F_E = eE = {1 \over 4\pi\epsilon_0} {Ne^2 \over R_E^2} }$ (away from the Earth) whereas the gravitational force on that same electron will be $\ds{ F_G = m_e g = G {M_E m_e \over R_e^2} }$ (toward the Earth). When we have as many ``extra'' electrons as possible, the two forces will just cancel (F_E = F_G), giving $\ds{ {1 \over 4\pi\epsilon_0} N e^2 = G M_E m_e }$ or $\ds{ N = {4\pi\epsilon_0 G M_E m_e \over e^2} = {4\pi\epsilon_0 g m_e R_E^2 \ . . . . . . 37 \times 10^6)^2 \over 8.99 \times 10^9 \times (1.602 \times 10^{-19})^2 } }$ or  (Note that this is only $2.52 \times 10^{-7}$ C!  Since it is possible to pick up as well as lose electrons from the solar wind, the Earth remains electrically neutral to uncanny precision!)