import ElectroMag;
real u = 1cm;
-real L = 3;
-real dy = 1;
+real L = 5;
+real dy = 2;
int n=3;
Charge q = aCharge((0,0)*u, q=1, L="$q$");
Wire top_plate = Wire((0,0.5*dy)*u, (L,.5*dy)*u);
Wire bottom_plate = Wire((0,-.5*dy)*u, (L,-.5*dy)*u);
-path p = q.center;
-real x;
-real y;
-for (int i=0; i<=100; i+= 1) {
- x = q.center.x+L*i/n;
- y = q.center.y+.4*dy*(x/L)**2;
- p = p..((x,y)*u);
-}
-draw(p, grey);
-
Vector E;
for (int i=0; i<n; i+=1) {
- E = EField((L*(i+.5)/n, -.4*dy)*u, mag=.8*dy*u, dir=90, L="$E$");
+ E = EField((L*(i+.5)/n, -.35*dy)*u, mag=.7*dy*u, dir=90, L="$E$");
E.draw();
}
+path p = q.center;
+real x, y, frac;
+int m = 20;
+for (int i=0; i<=m; i+= 1) {
+ frac = i/m;
+ x = L*frac;
+ y = 0.4*dy*frac**2;
+ p = p..(q.center + (x,y)*u);
+}
+draw(p, grey);
+
top_plate.draw();
bottom_plate.draw();
v.draw();
q.draw();
+
+draw_ijhat(q.center - (1,0)*u);
\end{asy}
\end{center}
From the forces on the drop in each direction
\Delta y &= \frac{qE}{2m} \Delta t^2 \\
q &= \frac{2m \Delta y}{E \Delta t^2}
= \frac{2\cdot1.4\E{-11}\U{kg}\cdot3.0\E{-4}\U{m}}{8.0\E{4}\U{N/C}\cdot(1\E{-3}\U{s})^2}
- = \ans{1.05\E{-13}\U{C}}
+ = \ans{1.0\E{-13}\U{C}} \;.
\end{align}
\end{solution}