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Chapter 26: Q. 6 (page 736)

An electron is released from rest atx=2m in the potential shown in Figure Q26.6. Does it move? If so, to the left or to the right? Explain.

Short Answer

Expert verified

The electron makes a Rightwardmovement.

Step by step solution

01

Given information and formula used    

Given :

Initial position of electron :x=2m

Graph -

Theory used :

Electric fieldis the negative slope (鈥渄erivative鈥� or 鈥済radient鈥�) of the potential.

E鈫�=-螖V螖x

Potential energy is related to electric potential by the charge. It only makes sense if you know the sign & magnitude of the test charge you want to use.

02

Determining the direction of the electron 

The electron makes a Rightwardmovement.

Because the electric field is the inverse of the electric potential's slope, the electric field will be nonzero and pointing to the left at x=2m.

As a result, the electron will experience a net force to the right and accelerate in that direction.

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Most popular questions from this chapter

The electric potential along the x-axis is V=100e-2xV, wherex is in meters. What is Ex at (a)role="math" localid="1649581712352" x=1.0m and (b) role="math" localid="1649581719374" x=2.0m?

The metal spheres are charged to +-300. Draw this figure on your paper, then draw a plausible contour map of the potential, showing and labeling the equipotential surfaces.

Figure Q26.10 shows a 3Vbattery with metal wires attached to each end. What are the potential differences 鈭�V12=V2-V1,鈭�V23=V3-V2,鈭�V34=V4-V3,and鈭�V41=V1-V4?

What is the potential difference 鈭�V34in Figure EX26.16?

a. Suppose that E鈫�=0鈫�V/mthroughout some region of space. Can you conclude that V=0Vin this region? Explain.

b. Suppose that V=0V throughout some region of space. Can you conclude that E鈫�=0鈫�V/m in this region? Explain.
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