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The surface charge density on an infinite charged plane is -2.0×10-6C/m2 . A proton is shot straight away from the plane at 2.0×106m/s. How far does the proton travel before reaching its turning point?

Short Answer

Expert verified

The proton travels18cm before reaching its turning point .

Step by step solution

01

Given information and formula used

Given :

The surface charge density on the infinite charged plane : -2.0×10-6C/m2

Proton shot straight away at the speed : 2.0×106m/s

Theory used :

The Electric field owing to an infinite plane is calculated as follows:

E=σ2ε0

Newton's law :

12m((vf)2(vi)2)=qEd

02

Calculating how far the proton travels before reaching its turning point 

Due to the action of the infinite plane's electric field, this proton will come to a halt before returning. We can compute the distance it goes before turning back using conservation of energy :

12m((vf)2(vi)2)=qEdd=m((vf)2(vi)2)2qE=m((vf)2(vi)2)2q(σ/2ε0)=(1.67x10-27kg)(0-(2.0x106m/s)2)(8.854x10-12C2/N·m2)(1.6x10-19C)(2.0x10-6C/m2)d=18cm

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Most popular questions from this chapter

The two parallel plates in FIGURE P23.53are 2.0cmapart and the electric field strength between them is 1.0×104N/C. An electron is launched at a 45 angle from the positive plate. What is the maximum initial speed v0 the electron can have without hitting the negative plate?

An electron in a vacuum chamber is fired with a speed of 8300km/stoward a large, uniformly charged plate 75cm away. The electron reaches a closest distance of 15cm before being repelled. What is the plate’s surface charge density?

A proton orbits a long charged wire, making 1.0×106 revolutions per second. The radius of the orbit is 1.0cm. What is the wire’s linear charge density?

The irregularly shaped area of charge in FIGURE Q23.7 has surface charge densityηi. Each dimension (x and y) of the area is reduced by a factor of 3.163.

a. What is the ratio ηf/ηi , where ηfis the final surface charge density?

b. An electron is very far from the area. What is the ratioFf/Fi of the electric force on the electron after the area is reduced to the force before the area was reduced?

A problem of practical interest is to make a beam of electrons turn a 90°corner. This can be done with the parallel-plate capacitor shown in FIGURE. An electron with kinetic energy 3.0×10-17Jenters through a small hole in the bottom plate of the capacitor.

a. Should the bottom plate be charged positive or negative relative to the top plate if you want the electron to turn to the right? Explain.

b. What strength electric field is needed if the electron is to emerge from an exit hole 1.0cmaway from the entrance hole, traveling at right angles to its original direction?

Hint: The difficulty of this problem depends on how you choose your coordinate system.

c. What minimum separation dminmust the capacitor plates have?

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