Figure 19-38 shows an electron entering a parallel-plate capacitor with a speed

Question

Figure 19-38 shows an electron entering a parallel-plate capacitor with a speed of v = 5.75 ? 106 m/s. The electric field of the capacitor has deflected the electron downward by a distance of d = 0.550 cm at the point where the electron exits the capacitor.

Figure 19-38 shows an electron entering a parallel-plate capacitor with a speed of v = 5.75 ? 106 m/s. The electric field of the capacitor has deflected the electron downward by a distance of d = 0.550 cm at the point where the electron exits the capacitor. Find the magnitude of the electric field in the capacitor. Find the speed of the electron when it exits the capacitor.

Explanation / Answer

F=ma=qE

a=q/mE

solve the equations of motion for a

x=vt

delta y(write the symbol delta)=1/2 at^2

=1/2 a(x/v)^2

=a(x^2/2v^2)

a=(2v^2)(deltaY)/x^2

=q/mE

solve for E

E=2mv^2deltay/qx^2

=2(9.11*10^-3kg)(5.24*10^6m/s)^2(0.00618m))/1.60*10^-19c)(0.0225m)^2

=4.13*10^3 N/c

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.