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Chapter 16: Q70P (page 670)

What is the electric potential at a location 2.5×10-9 mfrom an electron?

Short Answer

Expert verified

The value of electric potential is 0.57 V.

Step by step solution

01

Identification of given data

The given data can be listed below,

  • The location of electron is, r=2.5×10-9 m
02

Concept/Significance of electric potential

The effort done by the field in moving unit positive charge from the reference point to the place where potential is to be found determines electric potential at any point in the field.

03

Determination of the electric potential of electron

The potential of electron is given by,

V=Kqr

Here, K is the coulomb constant whose value is 9×109 N.m2/C2 , q is the charge on the electron whose value is -1.6×10-19C , and r is the distance between location and electron.

Substitute all the values in the above expression.

role="math" localid="1657098419012" V=9×109 N.m2/C2-1.6×10-19C2.5×10-9m=0.57V

Thus, the value of electric potential is 0.57 V.

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Most popular questions from this chapter

If throughout a particular region of space the potential can be expressed as V=4xz+2y-5z, what are the vector components of the electric field at location (x,y,z)?

In the region of space depicted in Figure 16.87 there are several stationary charged objects that are not shown in the diagram. Along a path A=B=C=D you measure the following potential differences:VB−VA=12 V ;VC−VB=−5 V ; VD−VC=−15 V. What is the potential differenceVA−VD ?

You move from location i at ⟨2,7,5⟩ mto locationf at ⟨5,6,12⟩ m. All along this path there is a nearly uniform electric field of ⟨1000,200,−510⟩ N/C. Calculate Vf−Vi, including sign and units.

A capacitor with a gap of 2 mm has a potential difference from one plate to the other of 30 V. What is the magnitude of the electric field between the plates?

long thin metal wire with radius rand lengthLis surrounded by a concentric long narrow metal tube of radius R, whereR>>L, as shown in Figure 16.86. Insulating spokes hold the wire in the center of the tube and prevent electrical contact between the wire and the tube. A variable power supply is connected to the device as shown. There is a charge+Qon the inner wire and a charge−Qon the outer tube. As we will see when we study Gauss’s law in a later chapter, the electric field inside the tube is contributed solely by the wire, and the field outside the wire is the same as though the wire were infinitely thin; the outer tube does not contribute as long as we are not near the ends of the tube. (a) In terms of the chargeQ, lengthL, inner radiusr, and outer radiusR , what is the potential differenceVtube−Vwire between the inner wire and the outer tube? Explain, and include checks on your answer. (b) The power-supply voltage is slowly increased until you see a glow in the air very near the inner wire. Calculate this power-supply voltage (give a numerical value), and explain your calculation. The lengthL=80 cm , the inner radiusr=0.7 mm, and the outer radiusR=3 cm. This device is called a “Geiger–Müller tube” and was one of the first electronic particle detectors. The voltage is set just below the threshold for making the air glow near the wire. A charged particle that passes near the center wire can trigger breakdown in the air, leading to a large current that can be easily measured.
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