}
void draw(picture pic=currentpicture) {
- frame picF;
+ picture picF;
+ picture picL;
+ label(picL, L);
+ pair pLabelSize = 1.2 * (max(picL)-min(picL));
path c = scale(radius)*unitcircle;
filldraw(picF, c, fill, outline);
- // TODO: label
+ label(pic = picF,
+ L = L,
+ position = (0,-(radius+pLabelSize.y/2)));
add(pic, picF, center);
}
}
return c;
}
+// neutral charge
+Charge neutralCharge(pair center=(0,0), real radius=2mm, pen outline=currentpen, Label L="")
+{
+ Charge c = Charge(center=center, q=0, radius=radius, outline=outline, L=L, fill=grey);
+ return c;
+}
+
// auto-signed charge
Charge aCharge(pair center=(0,0), real q=1, real radius=2mm, pen outline=currentpen, Label L="")
{
Charge c;
if (q > 0) {
c = pCharge(center, q, radius, outline, L);
- } else {
+ } else if (q < 0) {
c = nCharge(center, q, radius, outline, L);
+ } else {
+ c = neutralCharge(center, radius, outline, L);
}
return c;
}
\begin{problem*}{84}
Two tiny balls of mass $m$ carry equal but opposite charges of
-magnitude $q$. They are tied to the came ceiling hook by light
+magnitude $q$. They are tied to the same ceiling hook by light
strings of length $L$. When a horizontal uniform electric field $E$
is turned on, the balls hang with an angle $\theta$ between the
strings (Fig.~21.46). Assume that the force one ball exerts on the
electric field. \Part{a} Which ball (the right or the left) is
positive, and which is negative? \Part{b} Find the angle $\theta$
between the strings in terms of $E$, $q$, $m$, and $g$. \Part{c} As
-the electric field is gradually increased in strength, what does ytour
+the electric field is gradually increased in strength, what does your
result from \Part{b} give for the largest possible angle $\theta$?
\end{problem*}
+\begin{nosolution}
+\begin{center}
+\begin{asy}
+import Mechanics;
+import ElectroMag;
+
+real u = 1cm; // Length of 1 m on the page
+real L = 2; // Length of cable
+real phi = 30; // Half angle between cables
+real dy = .6; // Distance below charges to E field vector
+
+Charge a = neutralCharge(dir(-90+phi)*L*u, L="$q_R$");
+Charge b = neutralCharge(dir(-90-phi)*L*u, L="$q_L$");
+Wire La = Wire((0,0), a.center, L="$L$");
+Wire Lb = Wire(b.center, (0,0), L="$L$");
+Angle theta = Angle(a.center, (0,0), b.center, L="$\theta$");
+Surface s = Surface((a.center.x, 0), (b.center.x, 0));
+Vector E = EField(a.center - (0,dy)*u, mag=(a.center.x - b.center.x), dir=-180, L="$E$");
+
+s.draw();
+La.draw(rotateLabel=false);
+Lb.draw(rotateLabel=false);
+theta.draw();
+E.draw();
+a.draw();
+b.draw();
+\end{asy}
+\end{center}
+\end{nosolution}
+
\begin{solution}
+\Part{a}
+\begin{center}
+\begin{asy}
+import Mechanics;
+import ElectroMag;
+
+real u = 1cm; // Length of 1 m on the page
+real L = 2; // Length of cable
+real phi = 30; // Half angle between cables
+real dy = .6; // Distance below charges to E field vector
+
+Charge a = nCharge(dir(-90+phi)*L*u, L="$q_{R-}$");
+Charge b = pCharge(dir(-90-phi)*L*u, L="$q_{L+}$");
+Wire La = Wire((0,0), a.center, L="$L$");
+Wire Lb = Wire(b.center, (0,0), L="$L$");
+Angle theta = Angle(a.center, (0,0), b.center, L="$\theta$");
+Surface s = Surface((a.center.x, 0), (b.center.x, 0));
+Vector E = EField(a.center - (0,dy)*u, mag=(a.center.x - b.center.x), dir=-180, L="$E$");
+
+s.draw();
+La.draw(rotateLabel=false);
+Lb.draw(rotateLabel=false);
+theta.draw();
+E.draw();
+a.draw();
+b.draw();
+\end{asy}
+
+\begin{asy}
+import Mechanics;
+import ElectroMag;
+
+real phi = 30; // Half angle between cables
+
+Charge a = nCharge();
+Vector Fq = Force(a.center, mag=.4cm, dir=-180, L="$F_q$");
+Vector FT = Force(a.center, mag=1cm, dir=90+phi, L="$F_T$");
+Vector FE = Force(a.center, mag=Sin(phi)*FT.mag+Fq.mag, dir=0, L="$F_E$");
+Vector Fg = Force(a.center, mag=Cos(phi)*FT.mag, dir=-90, L="$F_g$");
+Fq.draw();
+FT.draw();
+FE.draw();
+Fg.draw();
+a.draw();
+
+Charge b = pCharge((-2cm, 0));
+Fq.center = FT.center = FE.center = Fg.center = b.center;
+Fq.dir += 180;
+FT.dir -= 2*phi;
+FE.dir += 180;
+Fq.draw();
+FT.draw();
+FE.draw();
+Fg.draw();
+b.draw();
+
+draw_ijhat((a.center+b.center)/2 - (0,1cm));
+\end{asy}
+\end{center}
+
+Because they have opposite charges, the two balls are attracted to one
+another by $F_q$. The combination of gravity $F_g$ and string tension
+$F_T$ also works to bring the balls back together. The only remaining
+force (which must push the charges apart to balance the previous
+attractive forces) is the force from the external electric field
+$F_E$. So $F_E$ on the right hand charge $q_R$ must be to the right,
+and $F_E$ on the left hand charge must be to the left. Because
+$\vect{F}_E=q\vect{E}$, this means that $q_R<0$ and $q_L>0$, so
+\ans{the left ball is positive and the right ball is negative}.
+
+\Part{b}
+Balancing forces on the right hand ball (taking care of sign up front)
+we have
+\begin{align}
+ \sum F_y &= F_T\cos(\theta/2) - F_g = 0 \\
+ F_t &= \frac{F_g}{\cos(\theta/2)} \\
+ \sum F_x &= F_E - F_q - F_T \sin(\theta/2) = F_E - F_q - F_g \tan(\theta/2) \approx F_E - F_g \tan(\theta/2) = 0 \\
+ F_E &= qE = F_g \tan(\theta/2) = mg \tan(\theta/2) \\
+ \theta &= \ans{2 \arctan\p({\frac{qE}{mg}})}
+\end{align}
+Where we used the fact (given in the problem) that $F_q \ll F_E$ to
+ignore $F_q$ in our answer.
+
+\Part{c}
+As $E\rightarrow\infty$, $F_E\rightarrow\infty$, so the balls get
+pulled further and further apart until $\ans{\theta=180\dg}$. In
+terms of our equation from \Part{b}, $\arctan(+\infty)=90\dg$, so
+$\theta=2\cdot90\dg=180\dg$.
\end{solution}
projects / course.git / commitdiff