QUESTION 1:
Can there be an electric field at a point where there is no charge? Can there be a charge at a place where there is no field?
QUESTION 2:
Let's say you are holding two tennis balls (one in each hand) and let's say these balls each have a charge Q. Estimate the maximum value of Q so that the balls do not repel each other so hard that you can't hold on to them.
Question 3:
An insulator is a material that ...
a) Cannot feel an electrical force.
b) Is not penetrated by electric fields.
c) Cannot carry an electric charge.
d) Two of the above choices are correct.
e) None the above choices are correct.
None of the students in the class chose response e, of 33 responses, 18 chose c (presuming a conventional use of the word "carry") and 15 chose d (presuming that insulators "block electricity" in a generic sense). None chose a, b, or e.
Below are the students responses.
I have used bold to indicate references to self fields.
I have used red to indicate a few other notable responses
Also, look for confusions among the terms "point," "charge," and "point charge," and for answers that invoke specific subatomic particles.
In the estimation question, note three levels of response:
QUESTION 2,The maximum value of Q would be very small a rough estimate might be around zero like 10^-12 C. Again the number would have to be very small.
QUESTION 3,d
Comments,
QUESTION 2,Assume the charges are held at a distance d and the maximum force that can be sustained by the hands F. And by Coulomb's law, the maximum charge is given by F=QQ/4piedd and Q= 2d SQRT(pie)
QUESTION 3,c
Comments,
QUESTION 2,Well, in order to repel the charges would have to be like (both either
positive or negative). This charge would have to be very small probably in
the nano coulombs, because 1 meter apart is not very far for the equation
F=k
QUESTION 1
Question 2/rxr.
QUESTION 3,c
Comments,
QUESTION 2,It the max. force you could hold was 5lbs. (49N), and holdin at a distance of 0.25m apart, a charge of greater than =1.31 e-5 C would be the maximum repulsion that you could overcome.
QUESTION 3,c
Comments,
QUESTION 2,Assume you hold them about1 meter apart and you are able to withstand a force of about 700N or approximately half your body weight. By Coulomb's law, Q is approximately 2.8 x 10(-4) C.
QUESTION 3,c
Comments,
QUESTION 2,Let's say that I could hold on with 980 N, and the distance at which I am holding the balls apart is 1 m. I would us the equation q = sqrt(F*r^2/k). Which, with all the numbers plugged in and calculated, q = 3.3*10^-4 C.
QUESTION 3,d
Comments,I had some trouble in the homework with the Electric field calculations (22-7). I wasn't really even sure where to begin even after reading the chapter.
QUESTION 2,Using the equation F=(kq^2)/(r^2) where F=500N, k=9.0*10^9, and r=.5m The charge can be upto 1.18*10^-4 coulombs or 118uC
QUESTION 3,c
Comments,
QUESTION 2,I'm guessing that I could hold together about 200N (I don't know if that's even reasonable or not) and I am choosing r to be a very small number (0.0000001m) since that if it were actually zero the math would fail. So my estimate is about 1.5*10^-11 C.
QUESTION 3,d
Comments,
QUESTION 2,I have to assume that the balls are about 0.5 meters across and that an average human can sustain a force of about 100 lbs. or 4.45E2 N. then 4.45E2 = (k)(Q^2/(.5)^2). and k=9E9 then Q = 1.11E-4.
QUESTION 3,c
Comments, I think an insulator can be penetrated by an electrical field, although I didn't find it in the book. I think this because they make voltage testers that test to see if you have electricity at an outlet ib your house. Of course some of these have metal tips, but some have plastic tips on them. The plastic ones must recognize an electrical field when brought close to the circuit, then light up with the help of a AA battery inside them.
QUESTION 2,I'm not sure how many Newtons I can hang on to but I'm going to say about 100. If the distance between the balls is about .3 meters then 100 times .3^2 gives me 9. If I divide this by k I get 1E-9. Now I take the sguare root of this to find Q which is about 30nC.
QUESTION 3,c
Comments,
QUESTION 2,I would have to find the forces exerted by the balls and estimate the force with which I could hold those two balls. F=K*(Q^2)/r^2 Let's assume that I can withstand a force of 200N holding the balls at a distnace of 20cm apart. The charge would be 2.9 X 10^-5. Any charge greater than that would force me to lose control of the balls.
QUESTION 3,d
Comments,
QUESTION 2,
Comments,
QUESTION 2,I took 1 nC= 10^-9C and then found how many newtons that is which is about 90N. I then divided by gravity to find the number of kilograms, 9.8kg, and that equals about 20 pounds, so I'm sure that it can be a bit more but not too much more.
QUESTION 3,d
Comments,
QUESTION 2,The charge Q would have to be a value such that the force exerted by each ball on the other would be approximately 500N. This divided by gravity would be about 50kg (assuming I could hold 50kg in each hand). Thus the charge Q would be approximately 4 x 10^-4 c. (also assumed was arm length of 1.75m)
QUESTION 3,c
Comments, A little confusing when asked to give estimates. Should I assume things such as r being my out-stretched arm length. I tried to work this out on paper just using the formula, but was not able. So I started assuming things about this problem to come up with an estimate.
QUESTION 2,If it were as great a one C then it would be pretty hefty. Thus it would have to be a fraction of a Coulomb but larger than a micro Columb. Just a guestimation of course.
QUESTION 3,d
Comments,Its amazing that an electron is so small, but when they get together, they get alot done. Sure wish we acted more like electrons sometimes.
QUESTION 2, If a person could hold around 40 kilograms with there hands 1 meter apart then the value of Q on each ball would be 2*10^-4 coulombs.
QUESTION 3,d
Comments,
QUESTION 2,Let's say I hold the balls 1 meter apart and I can only exert 450 N of force. The charge Q could be no more than about 2.25 e -4 C if I am able to hold onto them.
QUESTION 3,c
Comments,Before, in the past, I have at least seen the material once. So far this semester, things seem vaguely familiar. I hope things work out.
QUESTION 2,If my hands are approximately 20 cm apart and that I can hold approximately 100 lbs, then the value of Q would be approximately 2.2 x10^-5 C. (I'm doing this sans calculator.)
QUESTION 3,d
Comments,
QUESTION 2,Supposing that I can reasonably hold 50lbs: 50/9.8=5.1N 5.1N=(9e9)qq/16 q=9.1e-9
QUESTION 3,c
Comments,Can you clarify the idea of electrical fields. The description in the book is difficult to visualize.
QUESTION 2, I beleive the maximum value Q can be for me is about 1000C.
QUESTION 3,c
Comments,
QUESTION 2,My general guess is 1*10^-6 C. This guess is based on my being able to hold only ten newtons (me being very weak, maybe?) and trying to hold them about 20cm apart. In addition, I'm being very vague in my math.
QUESTION 3,d
Comments,
QUESTION 2, Human flesh is a good conductor, the balls would have a neutral if one were not wearing tennis shoes. Assuming that is not the case and that I could withstand 445 N (100lb), The maximum charge would be around 1 micro charge. F = (k * Q * Q) / square(r). Solving for Q gives Q = sqroot(F / k) * r = sqroot(445 / 8.988 * exp(9)) * 5 = 0.001112437284 C.
QUESTION 3,c
Comments,
QUESTION 2,The answer to this question depends on how strong or weak the person in question is and how far the t-balls are placed apart. Since I personally don't know how many newtons of force I am able to exert, I'm just going to assume that I can hold on to a force of, say 10 newtons, and the balls are 2 meters apart. Then, Q would equals 3.33e-4 coulomb.
QUESTION 3,c
Comments,
QUESTION 2,In this question the variables would depend on the persons ability. For me the max. would be around 50 kilograms for me and in this the radius at 1.0 x 10^9 gives a Q of about 52 C. This is only for this particular measurement.
QUESTION 3,d
Comments,I had a problem with number 2 i don't understand exactly how you wanted the question to be answered.
QUESTION 1b) What generates an electric field is a charge. If a charge is present at a point, then an electric field must be present at that point. The only exception is within a conductor, where the electric field is always zero within the conducting material.
QUESTION 2,
QUESTION 2a) Given that the tennis balls can be held at a distance of 47 inches and can be held with each arm at a force of 75lbs. The radius is 1.19 meters and the force is 333.62 newtons. The maximum charge that could be placed on each ball is 2.097x10-4 coulombs by calculation.
QUESTION 3,c
Comments,It has been displayed that electric force at small distances is far stronger than gravitational forces. Given the distance of electrons in close proximity within an atom, what keeps the electrons from crashing into the nucleus? What keeps the chaotic equlibrium that enables atoms from colapsing inward? Is it a nuclear force that repels electrons and attracts neutrons? What particles generate this force?
QUESTION 2,I don't know.
QUESTION 3,d
Comments,
QUESTION 2,For this question, two assumptions are made. A person can only exert a force of 600 Newtons on each side and the balls are one meter apart...By
using the formula
q=[(F * r^2)/k]^.5
then by substituting in, we get a value of a charge is
2.58e-4 C....
QUESTION 3,d
Comments,
QUESTION 2,Using Coulomb's law and estimating each hand can withstand about 44N at 1.3m apart. I get a charge on each ball of about +/- 9 X 10 -5C
QUESTION 3,c
Comments,
QUESTION 2,Assuming the maximum force that I can hold the balls is 700N (about 157 pounds) and I am holding them 1 meter apart, then the maximum charge Q could have would be 78 nC.
QUESTION 3,d
Comments,
QUESTION 2,For this problem I am estimating the maximum force I can hold to be 200 Newtons; in addition the distance I will try to maintain is 1 meter. Since force equals 'k', 8.988 x 10^9 N*m^2/C^2, multiplied by the charge squared (in this case) divided by the distance squared, the maximum charge I could handle would be 1.49 x10^-4 C.
QUESTION 3,d
Comments,
QUESTION 2,force=k(
QUESTION 1
QUESTION 2)/r2 r^ sqrt(r2/9*10to the 9th(-i^)=Q
QUESTION 3,c
Comments,I still am confused about field and force. Had trouble last night connecting to iupui to do warm up -- slow when working to connect when I did the puzzle. Probably because of all the rookies?
QUESTION 2,I figure that I could hold back about 1kg or about 10 newtons for each ball. (maybe more or maybe I'm crazy) This would be a net force of 20 newtons measured from either ball. I'll assume a distance of 1 meter apart for ease of calculation(I can be lazy) 20N= 9x10(ninth)(qsquared)/1 according to my calcuations, the max charge on each ball would be 4.71x10(-5)
QUESTION 3,d
Comments,for number 3, the choices are almost correct. According to my reading and general knowledge, something can be a great insulator but there is not a PERFECT insulator. That was the basis of my answer. Please correct me if I am wrong.
QUESTION 2,Based on working out at the gym, it is reasonable to expect be able to
counter a 10lb (44.48N) force easily.
Therefore, set F=44.48 Newtons
My hands strecthed out in front of me are aproximately .3 meters apart.
Therefore, set r=.3 meters
Using F=k(
QUESTION 1*
Question 2)/r^2 and solving for q, after plugging in the above, it would seem reasonable to expect: q=4.44*10^9 coulombs could be held in your hands. The sign of q of course would be opposite in order to satisfy the fact that the tennis balls are to repel each other.
QUESTION 3,c
Comments,I am particularly interested on the effects of electric fields and how they generate noise (extraneous signals) into instumentation systems and electronic control systems. For example, Data systems and electronic engine control systems that operate in the vincinity of high voltage ignition wires
QUESTION 2,I think I would know how to answer this, but I've run out of time. I am going to do this differently than last semester, but I'm not on track yet.
QUESTION 3,c
Comments,