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Chapter 23: Q. 15 (page 654)

Two10cmdiameter charged rings face each other,20cmapart. Both rings are charged to+20nC. What is the electric field strength at (a) the midpoint between the two rings and (b) the center of the left ring?

Short Answer

Expert verified

a.Electric field at midpoint between the two rings is0N/C.

b. Electric field at the center of the left ring is4.108脳103N/C.

Step by step solution

01

Calculation for Electric field at midpoint between the two rings (part a)

(a)

Diameter of both circle is10cm.

Radius is 5cm.

Charge on both circle is+20nC.

Distance between both circle is20cm.

Formula used:

Electric field due to a ring of radiusRand chargeQat a distanceris given by,

E=kQrr2+R23/2.......(1)

Electric field at the center of both rings will be0N/C.

Because electric field due to both rings will be equal and in opposite direction.

02

Figure for given situation

03

Calculation of electric field at the center of the left ring  (part b)

(b)

Electric field due left circle at its own center is0N/C.

This can find out by substituting0forrin equation(1).

The electric field generated by the right ring in the middle of the left circle is now in the left direction, and its magnitude may be measured using equation(1).

E=9脳109N路m2/C220脳10-9C20脳10-2m(0.2m)2+(0.05m)23/2

E=4.108脳103N/C

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Most popular questions from this chapter

Two 2.0-cmdiameter disks face each other, 1.0mmapart. They are charged to 卤10nC.

a. What is the electric field strength between the disks?

b. A proton is shot from the negative disk toward the positive disk. What launch speed must the proton have to just barely reach the positive disk?

The combustion of fossil fuels produces micron-sized particles of soot, one of the major components of air pollution. The terminal speeds of these particles are extremely small, so they remain suspended in air for very long periods of time. Furthermore, very small particles almost always acquire small amounts of charge from cosmic rays and various atmospheric effects, so their motion is influenced not only by gravity but also by the earth's weak electric field. Consider a small spherical particle of radius r, density 蚁, and charge q. A small sphere moving with speed v experiences a drag force Fdrag=6蟺畏rv, where 畏 is the viscosity of the air. (This differs from the drag force you learned in Chapter 6 because there we considered macroscopic rather than microscopic objects.)

a. A particle falling at its terminal speed vtermis in equilibrium with no net force. Write Newton's first law for this particle falling in the presence of a downward electric field of strength E, then solve to find an expression for vterm.

b. Soot is primarily carbon, and carbon in the form of graphite has a density of 2200kg/m3. In the absence of an electric field, what is the terminal speed in mm/s of a 1.0-渭m-diameter graphite particle? The viscosity of air at 20掳C is 1.8脳10-5kg/ms.

c. The earth's electric field is typically (150 N/C , downward). In this field, what is the terminal speed in mm/s of a 1.0 渭m-diameter graphite particle that has acquired 250 extra electrons?

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What are the strength and direction of the electric field at the position indicated by the dot in FIGURE EX23.1? Specify the direction as an angle above or below horizontal

What are the strength and direction of the electric field at the position indicated by the dot in FIGURE P23.35? Give your answer (a) in component form and (b) as a magnitude and angle measured cw or ccw (specify which) from the positive x-axis.
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