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In the hgure small, nonconducting ball of mass m 1.2mg and charge 2.8 x 10 8 C (distributed uniformly through its volume) hangs from an insulating thread that makes...

Question

In the hgure small, nonconducting ball of mass m 1.2mg and charge 2.8 x 10 8 C (distributed uniformly through its volume) hangs from an insulating thread that makes an angle 0 = 290 with vertical, uniformly charged nonconducting sheet (shown in cross section) Considering the gravitational force on the ball and assuming the sheet extends far vertically and into and out of the page, calculate the surface charge density. of the sheet:NumberUnits

In the hgure small, nonconducting ball of mass m 1.2mg and charge 2.8 x 10 8 C (distributed uniformly through its volume) hangs from an insulating thread that makes an angle 0 = 290 with vertical, uniformly charged nonconducting sheet (shown in cross section) Considering the gravitational force on the ball and assuming the sheet extends far vertically and into and out of the page, calculate the surface charge density. of the sheet: Number Units



Answers

SSM In Fig. 23-49, a small, nonconducting ball of mass $m=1.0 \mathrm{mg}$ and charge $q=2.0 \times$ $10^{-8} \mathrm{C}$ (distributed uniformly through its volume) hangs from an insulating thread that makes an angle $\theta=30^{\circ}$ with a vertical, uniformly charged nonconducting sheet (shown in cross section). Considering the gravitational force on the ball and assuming the sheet extends far vertically and into and out of the page, calculate the surface charge density $\sigma$ of the sheet.

For this problem. On the topic of gases law, we have shown a non conducting ball which has a mass of one mg in charge of 210% to the -8 columns. That is hanging from an insulating thread, which makes an angle of 30° with a vertical uniformly charged on conducting sheet. If we consider the gravitational force acting on the ball, we want to find the surface charge density sigma of the sheet. Now the forces acting on the ball, as shown in the diagram to the right and the gravitational force has a magnitude MG. Where M is the mass of the ball. The electrical force has magnitude Q. E. Where Q. Is the charge on the ball. And E. The magnitude of the electric field at the position of the ball. And the tension in the thread is denoted by T. The electric field produced by the plate is normal to the blatant points to the right. And since the ball is positively charged, the electric force on it also points to the right By the tension in the thread. Makes the angle 30°, which will call data with the vertical. And since the ball is in equilibrium then it falls on it vanishes. So the some of the horizontal components yields Q E minus T. Signed theater equal to zero. And then some of vertical components gives us the vertical component of the tension T co sign theater minus the weight of the ball. Mg. Again must equal to zero. Now we solve for the electric field E. And then continue sigma. Now the expression T Is equal to Q. E. Oversight leader from the first equation can be substituted into the second equation, which gives us Q. E equal to MG 10 data. And the electric field produced by a large uniform sheet of charge is given by E. Is equal to sigma over to absolutely not. Which means that few times the surface charge density sigma is over to absolutely not is equal to MG 10 data, which means that the surface charge density that we're looking for sigma is equal to two. Absolutely not. Mg 10. Dita Divided by Q. And if you put our values in we can find sigma. This is two times 8.85 times 10 to the minus 12. Cool. Um squared p newton meters squared Times the mass of the ball, which is one Times 10 to the -6 Kg Times acceleration due to gravity 9.8 m/km2 Times the 10 of 30°. Well divided by two Times 10 to the -8 columns. So we get the surface charges. Do you have the sheet to be five Times 10 to the -9 columns per square meter

Okay. See the problem three times for a variance of it. So you have a she with positive charges on it, And then you have, ah, charge to spend it on a string somehow. Ah, uh, I guess this is a good model for, like, pith balls, which I've seen in lab him anyway. And then, you know that this angle is 30 degrees, and then the goal is to get the charge density on the plates goes question mark, I'll go ahead and start it by writing out what's given. So, um, the masses. Um, so it's one milligram, and so I need to convert that to kilograms. And so that's one times 10 to the minus six in the charge is I think it's two ton of Muncie. Let me double check two point. Oh, times 10 to the minus six. Okay, that's also one point. Oh, um, I can write it. Okay, great. So do this. You want to just do a free body diagram on the so you have the weight down, which I'm gonna do no force of gravity. You have attention force. So call that have some tea, and then you have the repulsive electric force was gonna double check the signs of all these? Yes, positive. She and positive charge. And then the charge isn't moving. And so you know about the total for us zero. So, um, so then, yeah, you can say that the forces in either direction balance each other, so I'm gonna label this angle is 60 degrees that this one's 30 because you could make a 30 60 90 triangle. So in the ex direction, that means that the force from the electric field has to be balanced by the horizontal component of the tension forced. So that's f t cose 60. And then for why, you know that the, um, the vertical tension for us has to be balanced by gravity. So you can say f g equals f t times this sign of 60. And, um, basically, you have you two things you don't have are the electric force and the tension. But you can do some algebra to get the, uh, the eliminate tension because the problem doesn't ask about tension, though it is an unknown. And if you do that substitution, then hopefully you got what I got, which is f g over tan 60 equals f e. And then again, I used that they use is Trig. So signed 60 over CO 60 is 10 60. So now we it's time to plug in the gravitational force and the electric force in terms of, uh, the things that were given. So I'll go ahead and do that. And, um, after years, MGs lives mg over tan 60 equals a sigma over to Absalon night. So that's the electric forests. And then our excuse me, the electric field and then forces charged times. Um, Times Field. And so So you got that. And so algebraic Lee isolating sigma. I'll do that on the next page. It's probably not necessary. It feels more complete, though, if I do so it's to Absalon, not mg. I feel like I'm cutting out all the trig and you need help with that. Um then it's probably like, I don't know, this is a bit more basic than than all the trig, but yeah, it feels a lot more complete to write this out this way on DDE. Um, so if I plug in everything So, um JIA JIA's nine pointy If, uh, probably didn't forget that somehow it's a memorable number, I think. Um, yes. When I put that into my calculator, I got five point. Looks like a six 5.0. It was like a lag here. 5.0 times. 10 to the minus nine Newton per pool. Meter squared. I'm just gonna double check the calculation. Okay. Great. It looks good to me.

So we're going to essentially balance the horizontal forces. This would give us that t attention force times. Sign of data equals the charge Q Times, the electric field e. And we're going to then bounce the vertical forces. This is giving us t co sign of Fada equaling the weight mg and combining these two equations we have. That tangent of stada would be equaling the charge multiplied by the electric field divided by the weight mg. And so we can then say that fate of the angle would be equaling arc tanne of the charge multiplied by the electric field divided by M. G. And so we can then solve for the electric field. We know that the electric field for a sheet of charge would be the unit for Sigma the uniforms surface charged density divided by two times Absalon, not the primitive ity of free space. And so this would give us a 2.50 times 10 to the negative ninth cool arms per square meter. This would be divided by two times 8.85 times 10 to the negative 12th. Ah, cool arms squared current Newton per square meter. And this is giving us 1.41 times 10 to the second Newtons per Coolum and so theta would be equaling are 10 of the charge, 5.0 times 10 to the negative eighth. Cool ums multiplied by 1.41 times 10 to the second Newtons per Coolum And then this would be divided by the mass of 4.0 times 10 to the negative. Sixth kilograms multiplied by 9.80 meters per second squared and we find that the angle is equaling 10.2 degrees. This would be our final answer. That is the end of the solution. Thank you for one.

Hi in this given problem there is a uniform electric field in a horizontal direction from left. So right and as we know elective it always goes away from positive to negative. So this is positive and this is negative and in this electric field there is have all suspended with the help of a threat and the ball is given a negative charge. So under the influence of this electric field the ball is real estate from the vertical such that the thread makes an angle of going to disagree with the vertical. In this situation weight of the wall will be acting vertically downward but electrostatic force acting over it being negative, it will be attracted towards positivity. So this is electrostatic force which is given by charge. You time selected few. This will be the tension in that thread. So if this angle is 20 degree here this angle will also be 20 degree. So this tension will be resolved into two parts vertically upward. This will be T post 20 degree horizontally right word. This is p sign. 20 degrees charge given to the ball is negative. We have to find it now as the ball is in equilibrium. So there should be no net horizontal force and no net vertical force sitting on the ball. So it's creating the horizontal components He signed 20° is equal to Q into E. And he cause 20 degree vertically upwards. Could be equal to wait acting vertically downward. Now if you divide these two equations canceling tension we get and 20° is equal to U. E. Bye MG. We get an expression for that charge. Unknown charge which is U. Is equal to M. G. Times of 10 20 degree divided by electric field. So plugging in all the non values here, this charge will be given by 0.60 graham was the mass of the ball, so this is 0.60 into 10 days, par minus three kg, Multiplied by 9.8 acceleration to the gravity, multiplied by 1020° and divided by Electric field which has been given as 700 Newton park colon. So it comes out to be 3.1 in 2 10 for minus six colon. Or we can say this is 3.1 Micropal. Um and that is negative, which is the answer for this Given problem here. Thank you.


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