I have a positive and a negative charged plate running parallel to one another. To find the Electric Field to the left of the plates, I say:
E dot dA = Qenclosed / permittivity
Then E times 2pi(r) = Qenclosed / Permittivity
then E = Qenclosed / 2pi(r) x Permittivity
and since Qenclosed...
I'm asked to derive the mathematical expression for the erlationship of centripetal force as a function of T, m, and R.
I've found, from data, that F=0.615/T^2 = 0.05m = 1/4R
how would I bring it all together to form one equation involving T^2, m, and R?
Any help, thanks.
If Car A is going 30mi/hr east, and Car B is going 10mi/hr east. What is the relative velocity of Car A to Car B?
Wouldn't this just be Va - Vb = Vrelative? So.. 30mi/hr - 10mi/hr, the answer would be 20mi/hr?
Any help, thanks. :smile:
But the problem should not be this easy. Our teacher had said that it's not what you think it is.
Therefore, I had come up with this, which doesn't come out to make sense for some reason. Maybe someone could point out why?
Average Speed = Distance Traveled / Time it took to travel
You move from point 1 to point 2, and then from point 2 back to point1, using the identical path. if your average speed from point 1 to point 2 is S1 and the average speed from point 2 back to point 1 is S2. What's the average speed of the entire trip there and back, in terms of S1 and S2...
How would you go about finding a perpendicular vector, to two 3 dimensional vectors? One way, I solved is using the cross product of the two vectors. Splitting the i's j's and k's up and solving using a determinent. But, what's another way to do it?
OK another thing I've thought up is saying...
F = Y(delta-Laluminum / L-naught)A
F = Y(delta-Lsteel / L-naught)A
since force is equal throughout?
Then set the two equations equal to one another, where A cancels out?
Can anyone confirm this is a way of doing it?
If Aluminum and steel are connected as one rod, and are attached to the wall and is pulled upon. How would the change in length for each material of the rod be calculated? I'm given that the Aluminum section is twice as long as the steel section, and the total change in length of the whole rod...
The ditto the teacher handed out to us was not his. A previous teacher had made it, and people were asking about it today, and for him to help us with it. He put it on the board, looked at it, and said - forget it, omit it. although he said he may give extra points for it solved.
Hurkyl - I...
Okay, I'm in Precalculus Honors, and have not learned about Derivatives yet, and this is an extra credit problem.
I've gathered that you must find the derative of the curve at the point of tangency, and that is the slope of the line. I'll google some more, but any help?
If 15eV photon interacts with a Hydrogen atom at groundstate [-13.6eV], and all 15eV is transferred to the atom. How would the KE of the ejected electron be found? and what is the de Broglie wavelength of the electron?
For the Kinetic Energy, i said 15eV = KE + 13.6eV, KE = 1.4eV
If there's a loop exiting a B-field, where half of it is in the b-field and half of it is out. How would a potential difference be found in the wire?
I said that V = BLv, (where L is the height of the loop, disregarding the width) and found V to be 3v. Although, there is one resistor on the...
The question runs me through a series of questions actually, and I think they are all linked somehow.
The first one asks what the sign of the charge of the particle is. And the answer I got was Negative, since the B-Field is going INTO the paper, and the velocity is to the right. Therefore...
Well I'm given two plates and a B-field. a charge is placed inbetween the two plates and moves paralllel to the two plates until the end, where it then makes a downward circle turn. All the while, the charge is in a B-Field.
the distance between the two plates is given
the B-field is given...
OHHH wow. Thanks a lot.
Ok, so for the net electric field to be 0, kq1 / (1+r)^2 + kq2/(r^2) = 0
1) k's drop out, leaving q1 / (1+r)^2 + q2/(r^2) = 0
2) (6.8x10^-6) / (1+2r+r^2) + (-1.7x10^-6) / (r^2) = 0
- multiply through by (1+2r+r^2)(r^2)
3) (6.8x10^-6)(r^2) +...
The definition of electric field strength? The amount of force that is pulling or pushing on charges surrounding a charge?
E=F/q, so the Electric Field strength would be the Force [kq1q2/r^2] divided by the charge, q.
That means, E = kq1 / r^2.
Oh, also - Eleectric...
OHhhhh is it:
Kq1 / r-from-q1 + kq2 / r-from-q2 = 0
so the K's cancel out leaving:
q1 / r1 + q2 / r2 = 0
q1 and q2 are knowns but r1 and r2 are unknowns. how would that be solved?
Would it be, kq1qe / r1^2 = kq2qe / r2^2
where q1 = q1
q2 = q2, and qe = equilbrium charge...
So what I did was:
Kq1/(1+r) + Kq2/r = 0, making r =the distance away from q2 outside of the system to be the equilibrium point. so q1 is 1m + r distance away from the point, and q2 = r distance away from the point.
K's cancel out, you get:
q1/(1+r) + q2/r = 0
Some algebra, turns out r =...
I thought W = q(Vb - Va)? Or is that only valid when it's one charge moving from one place to another, and W = qVtotal, when there's more than one charge?
And yeah, I think I've fallen behind because I can fully grasp the idea of F=k(q1)(q2) / r^2, and E = F/q, but I don't fully understand...
The way to solve this, i tried is by saying F = k(q1)(q2) / r^2. So, the force between the two charges is (9x10^9)(2.6x10^-8)(5.5x10^-8) / (1.4^2). Although, from there I'm not sure where to go because that just solves for the force inbetween the two charges and not the EPE or even more...
I got it. Thank you.
I did F = ma = 0.
and one sphere = equilibrium, created a right triangle with mg, Tension, and Electrical force. Knowing [mg] and the angles, I solved for the Electrical Force.
Then set that = to (k)(2q)(q) / r^2, where r = 0.0523 and k = 9x10^9.
q = 1.96 x...
So finding exactly whether it's both positive or negative can not be done?
Also, would my equations look like this?
er... would the vectors added be: m1g, m2g, k(2q)(q) / r^2, k(2q)(q) / r^2, T1, and T2?
add them as vectors and set them equal to zero and solve for q?
First, I said that the distance between the two spheres is 0.0523m since you can form a right triangle using 0.5m [50cm] as the hypotonose, and 3 degrees as an angle. and multiplying that answer by 2 giving me the distance between the two spheres.
From that I know one sphere = 2q, while the...
Can anyone help me with number two please? Any other suggestions?
Potential difference between the block's A and B position is 90 correct?
since 1/2mv^2 = mgh
becomes 1/2v^2 = gh
v = 42, g = 9.8, that solves for h = 90. so potential difference = 90?
Can anyone tell me if that's...
OR you could draw a right triangle, make one angle theta, and put in 4 as the side opposite theta, and 7 as the hypotonose (since Sin = opposite/hypotonose = 4/7 [change -4 to 4 since you can't have a negative length of a side])
Use pythagorem theorm to find that the 3rd side = radical 33...
sin(theta) = -4/7, cos(theta) = unknown.
but you know:
sin^2(theta) + cos^2(theta) = 1
(-4/7)^2 + cos^2(theta) = 1
solve for Cos, and since it's a squareroot you're going to get a positive and negative answer, take the positive answer (since cos's positive in the restrictions given of...
For number 2, I thought it was two situations, one with a charge and another with a mass?
When you say qV = 1/2mv^2, that's for the charge - what would the mass equal to?
I'm a bit confused, could you explain a little bit more please?
(The other two problems I figured out with your...
Hey - just asking for some help on some electric potential energy questions.
First I used the equation 1/2mv^2 = q(Va-Vb)
so v = sqrt (((2)(25000)(q)) / m )
and Va-Vb = W / q, so q = W/(Va-Vb)
but then I'm stuck and don't know where to go from there since the mass and the charge (q)...
So the question reads:
"A small drop of water is suspended motionless in air by a uniform electric field that is directed upward and has a magnitude of 8480 N/C. The mass of the water drop is 3.50 x 10^-9 kg. (a) Is the excess charge on the water drop positive or negative? Why? (b) How many...
Electric Field question:
Two charges are located on the x axis: q1 = +6.0 micro-Coloumbs at x1=+4cm, and q2=+6 micro-coloumbs at x2 = -4cm. Two other charges are located on the y-axis: q3 = +3 micro-Coloumbs at y3=+5cm, and q4 = -8 micro coloumbs at y4 = +7cm. Find the net electric field...
Thanks for the response.
Ohh okay, so after
A and B are touched, it becomes: A=+2q, B=+2q, C=0q
then C and A touch, becoming: A=+1q, B=+2q, C=+1q
then C and B touch, resulting: A=+1q, B=+1.5q, C=+1.5q
Is that how it should be done?
The question goes as follows:
The correct answers are:
Although, my logic doesn't seem to give me these answers, I was wondering if anyone can explain why and how to approach this problem.
My logic goes as follows:
Sphere A = +5q
Sphere B = -q