In Fig. 20.22, assume that the charged particle is a proton of charge q = +e. The proton is released from rest at the positive plate. In this case, each of the two oppositely charged plates which are d = 2 cm apart has a charge per unit area of σ = 5μC/m². (a) What is the magnitude of the electric

Question

In Fig. 20.22, assume that the charged particle is a proton of charge q = +e. The proton is released from rest at the positive plate. In this case, each of the two oppositely charged plates which are d = 2 cm apart has a charge per unit area of σ = 5μC/m². (a) What is the magnitude of the electric field between the two plates? (b) What is the speed of the proton as it strikes the second plate?

Accepted Answer

(a) The electric field arises from two infinite plates, Thus:

[latex]E={\frac{\sigma}{2\epsilon_{o}}}+{\frac{\sigma}{2\epsilon_{o}}}={\frac{\sigma}{\epsilon_{o}}}={\frac{5 imes10^{-6}\,{\mathrm{C}}/{\mathrm{m}}^{2}}{8.85 imes10^{-12}\,{\mathrm{C/N}}.{\mathrm{m}}^{2}}}=5.65 imes10^{5}\,{\mathrm{N/C}}[/latex]

(b) We first find the proton’s acceleration from Newton’s second law:

[latex]a={\frac{F}{m}}={\frac{e E}{m}}={\frac{(1.6 imes10^{-19}\,{\mathrm{C}})(5.65 imes10^{5}\,{\mathrm{N}})}{1.67 imes10^{-27}\,{\mathrm{kg}}}}=5.41 imes10^{13}\,{\mathrm{m/s}}^{2}[/latex]

Then, using x = [latex]v_{0} t + \frac{1}{2} a t^{2}[/latex], we find that d = [latex]\frac{1}{2} a t^{2}[/latex]. Thus:

[latex]t=\sqrt{\frac{2d}{a}}=\sqrt{\frac{2(0.02 \ {\mathrm{m}})}{5.41 imes10^{13}\,{\mathrm{m}}/s^{2}}}=2.72 imes10^{-8}\,{\mathrm{s}}[/latex]

Finally, we use v = [latex]v_{0}[/latex] + a t to find the speed of the proton as follows:

[latex]v=a\,t=(5.41 imes10^{13}\,\mathrm{m}/s^{2})(2.72 imes10^{-8}\,\mathrm{s})=1.47 imes10^{6}\,\mathrm{m}/s[/latex]

Question 20.6: In Fig. 20.22, assume that the charged particle is a proton ......

Related Answered Questions

An infinite sheet of charge is lying on the xy-plane as shown in Fig. 20.16. A positive charge is distributed uniformly over the plane of the sheet with a charge per unit area σ. Calculate the electric field at a point P located a distance a from the plane. ...

Verified Answer:

Four point charges q1 = q2 = Q and q3 = q4 = −Q, where Q = √2 μC, are placed at the four corners of a square of side a = 0.4 m, see Fig. 20.5a. Find the electric field at the center P of the square. ...

Verified Answer:

(Electric Dipole) Consider two point charges q1 = −24 nC and q2 = +24 nC that are 10 cm apart, forming an electric dipole, see Fig. 20.6. Calculate the electric field due to the two charges at points a, b, and c. ...

Verified Answer:

(The Dipole Field Along the Dipole Axis) A proton and an electron separated by 2 × 10^−10 m form an electric dipole, see Fig. 20.10. Use exact and approximate formulae to calculate the electric field on the x-axis at a distance 20 × 10^−10 m to the right of the dipole’s center. ...

Verified Answer:

Verified Answer:

In Fig. 20.23, assume that the horizontal length L of the plates is 5 cm, and assume that the separation D between the plates and the screen is 50 cm. If the uniform electric field has E = 250 N/C, and the electron’s initial speed v0 is 2 × 10^6 m/s, then; (a) What is the acceleration of the ...

Verified Answer: