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Problem 20.10Falling raindrops frequently develop electric charges. Does this create noticeable forces between the droplets? Suppose two 1.8 mng drops each have cha...

Question

Problem 20.10Falling raindrops frequently develop electric charges. Does this create noticeable forces between the droplets? Suppose two 1.8 mng drops each have charge of +25 pC , The centers of the droplets are at the same height and 0.44 cm apart:SubmitPrevious_AnswergCorrectPart BWhat horizontal acceleration doos this force produce on the droplets? Express your answer with the appropriate units0.195SubmitPrevious_Answerg Request AnswerIncorrect; Try Again; 3 attempts remaining

Problem 20.10 Falling raindrops frequently develop electric charges. Does this create noticeable forces between the droplets? Suppose two 1.8 mng drops each have charge of +25 pC , The centers of the droplets are at the same height and 0.44 cm apart: Submit Previous_Answerg Correct Part B What horizontal acceleration doos this force produce on the droplets? Express your answer with the appropriate units 0.195 Submit Previous_Answerg Request Answer Incorrect; Try Again; 3 attempts remaining



Answers

Falling raindrops frequently develop electric charges. Does this create noticeable forces between the droplets? Suppose two $1.8 \mathrm{mg}$ drops each have a charge of $+25 \mathrm{pC} ;$ these are typical values. The centers of the droplets are at the same height and $0.40 \mathrm{cm}$ apart. What is the approximate electric force between them? What horizontal acceleration does this force produce on the droplets?

Hi, Obr. Even this is the problem based on force experienced by yeah, charged particle and uniforms electric quit. And uh huh. Gravitational field. Uh huh. Here it is, given a real drop. Having that I omitted 2.5 millimeters. Okay. Falling through the air. Acquired a charge off 15 people. Cool. Um, if the uniforms electric field off 100 Newtons per column directed downward, right, then find the ratio off. Electrostatic forces and gravitational force on the drop. Yeah, his No, to see it. Yeah. Mass of the drug will be or by three pipe our Cuban to root here ruins that Anstey off water, which is 1000 k g permitted you. So the ratio off electrostatic force toe gravitational force will be Q E upon MGI e m g is for by three pi r Q role and to G, So the ratio can vary. The nets three Q upon food by hard to ruin to G substitute the value charges given 15 people Cool. Electric field is 100 right? 53.14 reduces 1.1 to 5 and tow Trento The par minus two and density is 1000. Yeah, this is my district millimeter is there. We have to convert it. So on solving it, the ratio will be 2.5 months. Six in tow, 10 to the power seven, that's all. Thanks for watching it.

Hi, everyone. This is the problem Based on force on the charge Particle do tube electric field and gravitational people Here it is given there is a raindrop having that I omitted. Two millimeters carrying the church. PICO Hula is in direct Lichfield off 15 in tow 10 to the power three Nuclear purple Um, we have good. Fine. The ratio off Electrostatic force sent gravitational force on the drop. So Bible major the mass of the drop which will be for volume into dense tree that is four by three Fire Cuban to row So ratio off electrostatic forces and gravitational force Electrostatic forces queuing to e gravitational forces MGI that is you into eat upon four by three pi r q ruin to G So this ratio will be 32 into eight upon food fight Aren't you ruin to G substituting the value church is given 12 in tow, 10 to the power minus but electric village 15 and to 10 to the power three by is 3.14 radius is one in tow, 10 to the power minus tree, right G nine Sorry. Density is how's it so on solving it? This ratio would be 4.387 That's not thanks for watching it

Well the magnitude of the electric force on the rim. The raindrop is, uh, found by using APP vehicles king times e right, and by substituting values F vehicles, our eight multiplied by 1.6. Multiply by 10 to the power minus 19 multiplied by two. Multiply by 10 to the power six and so force is equal to 2.6. Multiply by 10 to the power minus 12 Newton. Since the Charter of the raindrop is negative, electric force on it is directed downwards opposite to the direction of the electric field, so it is directed downwards directed downwards.

What are the mass and radius of the drop? So we have at terminal speed. M. G is equal to B. V. And we have em is equal to 4/3. Hi a cubed row of the oil B. is he going to six pi? So we get that a squared is equal to 18/4. You she T over a row of the oil G. And they gave us V. Is five times 10 to the third m over 20 seconds. We have the viscosity and we have the specific gravity. So we get that A. Is equal to 1.66 10 to the negative 3rd mm. So M is equal to four pi Times. 1.66 times 10 to the -6 meters. Cute Times 750 kg per meter cubed Oliver three. And we get 1.44 Times 10 to the 14th kilograms part B. If the droplet carries two units of electric charge and is in the electric field of 2.5 times into the fifth fold per meter. What is the ratio of the electric force to the gravitational force? Okay, so electric force to gravitational force we have, The electric force is Q. E. The gravitational force is MG. So Plugging in values we get 0.57.


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