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As shown in Figure 16.72, three large, thin, uniformly charged plates are arranged so that there are two adjacent regions of uniform electric field. The origin is at the center of the central plate. Location A is <-0.4,0,0>m, and location B is<0.2,0,0>m . The electric fieldE1 has the value <725,0,0>V/m, and E2is <-425,0,0>V/m.

(d) What is the minimum kinetic energy the electron must have at location A in order to ensure that it reaches location B?

Short Answer

Expert verified

(d) The minimum kinetic energy of the electron is -328×10-19J.

Step by step solution

01

Write the given data from the question.

The location of A,xA=-0.4,0,0m

The location of B, xB=0.2,0,0m

The electrical field, E1=725,0,0V/m

The electrical field, E2=-425,0,0V/m

The charge of the electron, q=1.6×10-19C.

02

Determine the formulas to calculate the change in the electrical potential, change in the kinetic energy and minimum kinetic energy.

The expression to calculate the change in the potential is given as follows.

ΔV=ExΔx+EyΔy+EzΔz

Here, Exis the electrical field inx- direction, Δxis the change in the distancex-direction,Ey is the electrical field iny- direction,Δy is the change in the distancey- direction,Ez is the electrical field in z-direction, Δzand is the change in the distance z-direction.

The expression to change in the kinetic energy is given as follows.

ΔK=qΔV …… (i)

03

Calculate the minimum kinetic energy of the electron.

(d)

The electron should have the enough kinetic energy to traverse the potential difference between A and B. therefore the minimum kinetic energy of the electron should be-328×10-19J ,So the electron can reach at the location B.

Hence the minimum kinetic energy of the electron is -328×10-19J.

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Most popular questions from this chapter

If the kinetic energy of an electron is 4.4×10-18J, what is the speed of the electron? You can use the approximate (non relativistic) equation here.

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