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Im stuck in a couple electromagnetics problems 
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#1
Feb1205, 07:17 PM

P: 177

i have been at this time since 2 this afternoon, and ive only got 3 of 8 problems completely done.
i need to get these done before i go insane. heres the first: plane x+2y=5 carries charge Ps = 6nC/m^2. determine E at (1,0,1) okay, this is obviously a surface charge problem. and E = integral of Ps*ds. i just solve for E than plug in the (1,0,1) right? what i cant figure out are the limits of integration. would this limit work for y? y = (5x)/2. that would be the upper limit and the lower limit would be 0. what would work for x? 


#2
Feb1205, 07:30 PM

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#3
Feb1205, 07:38 PM

P: 177

ahh, youre right. heres what i need to do than: E = integral of Ps*ds*unit vector /4*pi*Eo*r^2
unit vector = (1,0,1)  ........im not sure if i should just put (x,y,0) for the second part or not. 


#4
Feb1205, 07:43 PM

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Im stuck in a couple electromagnetics problems
Something that just occurred to me: What is the electric field of an infinite plane of charge? Does it depend on position?



#5
Feb1205, 07:52 PM

P: 177

according to my book, the electric field on the z=0 plane is Ps/2Eo * the z unit vector.
is x+2y = 5, that sounds like a line than a plane. 


#6
Feb1205, 08:15 PM

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About the first sentence you wrote: that result is true for any infinite plane (easiest derivation uses Gauss's law). The only difference is that for a general plane, the direcition of the field would be [itex] \hat{n} [/itex], where [itex] \hat{n} [/itex] is a unit vector normal to the plane.
About the second sentence: The equation describes a line in twospace. But your original statement of the problem called it a plane. The equation describes a plane in threespace. And obviously you are working in three space because all of your points have been given with three coordinates. So it is a plane. Visualise it this way. There is no z in the equation, so there are no restrictions on the zcoordinates that points in the the plane can have. At z = 0, the plane intersects the xyplane in a line: y = x/2 + 5/2. At z = anything, the same thing is true. So the plane in question is the one swept out by the line: y = x/2 + 5/2 as you move it up and down the z axis for all values of z. 


#7
Feb1205, 08:34 PM

P: 177

so the main idea of the problem is to find the the normal unit vector than right.
how would i do that? 


#8
Feb1305, 04:03 PM

P: 177

i figured out most of the other ones, but this one is still giving me some trouble. i just need to be pointed in the right direction. can anybody help me out?



#9
Feb1305, 08:16 PM

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Oops, sorry, yeah, heh. Finding the unit normal vector. A plane is completely determined by a point in space P_{0} that lies in the plane and a vector [itex] \mathbf{n} [/itex] that is orthogonal to the plane. (Can you see why?) That's the normal vector. If P_{0} has a position vector [itex] \mathbf{r_0} [/itex] and some other arbitrary point in the plane P = (x,y,z) has a position vector [itex] \mathbf{r} [/itex], then the vector given by [itex] \mathbf{rr_0} [/itex], which is the directed line segment between P_{0} and P, lies in the plane. This is true for any arbitrary point P in the plane, so as long as [itex] \mathbf{n} [/itex] is orthogonal to [itex] \mathbf{rr_0} [/itex], then it is perpendicular to the plane. In other words their dot product is zero:
[tex] \mathbf{n} \cdot (\mathbf{rr_0}) = 0 [/tex] is the vector equation of a plane. Turn it into a scalar equation. Let: [tex] \mathbf{n} = <a, b, c> [/tex] [tex] \mathbf{r} = <x, y, z> [/tex] P_{0} (our point lying in the plane) is given by: [tex] \mathbf{r_0} = <x_0, y_0, z_0> [/tex] And so that vector equation becomes: [tex] <a, b, c> \cdot <xx_0, yy_0, zz_0> = 0 [/tex] [tex] a(xx_0) + b(yy_0) + c(zz_0) = 0 [/tex] This is the scalar eqn of a plane through a point P_{0}(x_{0},y_{0},z_{0}) with normal vector n Your P_{0} is any point lying on the plane given in your problem, such as (1,2,0) If you collect terms in the last equation, you get ax + by + cz  ax_{0}  by_{0}  cz_{0} = 0 Call this equation [1] compare this with the actual equation of the plane: x + 2y 5 = 0 So we want a = 1, b= 2, c = 0. Does it work? Substitute into [1] (1)x + (2)y + (0)z  1(1)  2(2)  0(0) = 0 x + 2y 1  4 = 0 x + 2y 5 = 0. It works! So the normal vector given by <a,b,c> is just <1,2,0>. What do you notice? The x, y, and z components of the normal vector are just the coefficients of the x, y and z terms of the original plane equation! So although we solved this the long way, deriving everything, all you had to do was look at the coefficients of the x, y, and z terms of the plane equation, and note that they were 1, 2, and 0. This is because the plane equation is always in the form ax + by + cz + d = 0, just as we derived, and n = <a,b,c>. Now that you know the normal vector, I don't think you should have any trouble turning it into a unit normal vector. 


#10
Feb212, 11:41 AM

P: 3

You first need to understand that the plane containing the charges is an infinite plane.construct a gaussian pillbox and determine the direction of electric field.the pillbox is a right circular cylinder with it's axis perpendicular to the required plane.By symmetry,the field lines to this infinite plane body is only perpendicular to the plane.Now that you have determined the direction of the field,apply gauss's law
∫E.DA= Q(enclosed)/ε hence the field only due to the circular portions matter...Hence 2*E*A= Q(enclosed)/ε.... sigma is surface charge density 2*E*A = σ*A/ε implies E= σ/(2*ε)→1 this shows that for an infinite plane,the field is uniform IRRESPECTIVE OF THE DISTANCE FROM THE PLANE the given equation of the planes is x+2y=5 this implies that the direction ratios of the normal is <1,2,0>....hence the unit normal vector is (i+2j)/√5 →2...implies that the field is magnitude(given by 1) * unit vector(given by 2)..... 


#11
Feb212, 11:44 AM

P: 3

The unit vector is (i2j)/(sqrt(5)....the negative direction as the point given is on the same side as origin and direction of filed lines reverse...if only i had the ability to post a diagram on these threads :(



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