tony873004
Jun29-06, 03:20 PM
Figure 19-38 shows an electron entering a parallel-plate capacitor with a speed of v = 5.25 106 m/s. The electric field of the capacitor has deflected the electron downward by a distance of d = 0.626 cm at the point where the electron exits the capacitor.
http://www.webassign.net/walker/19-38alt.gif
(a) Find the magnitude of the electric field in the capacitor.
b) Find the speed of the electron when it exits the capacitor.
Part A:
E=F/q, but q is not given
F=ma, but m is not given
F=kQq/r2
I don't know where to start.
Part B:
Compute time:
\begin{array}{l}
t = \frac{d}{v} \\
\\
t = \frac{{0.025m}}{{5.25 \times 10^6 m/s}} = 4.2857 \times 10^{ - 8} s \\
\end{array}
Compute acceleration:
\begin{array}{l}
y_f = y_i + v_i t + 0.5at^2 \\
\\
a = \frac{{y_f - y_i - v_i t}}{{0.5t^2 }} \\
\\
a = \frac{{0.626m - 0 - 0t}}{{0.5t^2 }} \\
\\
a = \frac{{0.00626m}}{{0.5\left( {4.2857 \times 10^{ - 8} s} \right)^2 }} = 6.81644 \times 10^{12} m/s^2 \\
\end{array}
Compute velocity:
\begin{array}{l}
v_{y,f} = v_{y,i} + at \\
\\
v_{y,f} = 0 + 6.81644 \times 10^{12} \times 4.2857 \times 10^{ - 8} \\
\\
v_{y,f} = 292133.33 \\
\\
v_f = \sqrt {v_{y,f}^2 + v_x^2 } \\
\\
v_f = \sqrt {(292133.33m/s)^2 + (5.25 \times 10^6 m/s)^2} \\
\\
v_f = 5.258122 \times 10^6 m/s \\
\end{array}
But this answer is wrong.
http://www.webassign.net/walker/19-38alt.gif
(a) Find the magnitude of the electric field in the capacitor.
b) Find the speed of the electron when it exits the capacitor.
Part A:
E=F/q, but q is not given
F=ma, but m is not given
F=kQq/r2
I don't know where to start.
Part B:
Compute time:
\begin{array}{l}
t = \frac{d}{v} \\
\\
t = \frac{{0.025m}}{{5.25 \times 10^6 m/s}} = 4.2857 \times 10^{ - 8} s \\
\end{array}
Compute acceleration:
\begin{array}{l}
y_f = y_i + v_i t + 0.5at^2 \\
\\
a = \frac{{y_f - y_i - v_i t}}{{0.5t^2 }} \\
\\
a = \frac{{0.626m - 0 - 0t}}{{0.5t^2 }} \\
\\
a = \frac{{0.00626m}}{{0.5\left( {4.2857 \times 10^{ - 8} s} \right)^2 }} = 6.81644 \times 10^{12} m/s^2 \\
\end{array}
Compute velocity:
\begin{array}{l}
v_{y,f} = v_{y,i} + at \\
\\
v_{y,f} = 0 + 6.81644 \times 10^{12} \times 4.2857 \times 10^{ - 8} \\
\\
v_{y,f} = 292133.33 \\
\\
v_f = \sqrt {v_{y,f}^2 + v_x^2 } \\
\\
v_f = \sqrt {(292133.33m/s)^2 + (5.25 \times 10^6 m/s)^2} \\
\\
v_f = 5.258122 \times 10^6 m/s \\
\end{array}
But this answer is wrong.