I have an easy question which I've been thinking about for a while..
Lets say I want to take the derivative of a function y = f(x) with respect to x, we would get.
dy/dx = f'(x).
In the couple of books I've skimmed through, they all say that dy/dx is not a ratio but the notation that...
Sorry, I left all the integration out..
This is what I did.
General Eq. for Eline = pL/(2piε0r),
1.r is the distance from the line and
2. pL is the charge density of the line
The given the problem.
Eline = ELeft Line + ERight Line
= pL/(2piε0)[1/(d/2 - x) + 1/(d/2+x)] { points in the...
Am I doing your approach correctly?
Actually the work in the y direction should be due to boh the line charges and plate... I didn't include the line charges.
Ok well the work in moving charge q is
q(pL/(piε0)ln[(d-a)/a])
The total work in moving charge q from the right wire to the left then to the plate is...
q{pL/(piε0)ln[(d-a)/a] + psd/(2ε0)} = qΔV
pL = QL/L
ps = Qs/A ; A = L2
ΔV(x) = pL/(piε0)ln[(d-a)/a] = Change in potential from the surface of
pL>0 to pL< 0 wire.
pL: Units C/m
i know you asked for the work but is it ok if i just calculate the potential?
I need guidance calculating the capacitance between two infinitely parallel long wires that are parallel to a large flat metal sheet.
Condition:
V = 0 at the origin.
Both plane and lines have uniform charge.
Left line: Negative charge density
Right Line: Positive charge density...
The problem and solution are in the document.
My question is in regards to the solution about calculating the charge on the capacitor. Would an equally correct solution for the charge be...
Let σs be the surface charge density on the plate.
The plate is a conductor and I would assume that the...
Just another though,
considering the surface of a conductor, we know the equipotential surfaces very neer the surface of the conductor is neerly the shape of that conductor.
Therefore, by knowing the equation of the surface we also know the equation of the equapotential lines neer the surface of...
I know the E-Filed at a point on the surface of the cone is perpendicular to the surface of the cone. Therefore the surface of the cone is an equipotential surface.
In spherical coordinates we have (R,θ,ø)
R is the distance from the origin
θ is the angle from the positive z-axis
ø is the...
In the example in the attachment, Laplaces Equation is used to find the potential of a cone.
My qustion is, How do they know the potential only varies with angle theta (theta is the angle between the positive Z-axis and the surface of the cone.)
In my book is says the electric flux density D is equal to εE if the medium is linear and isotropic, where E is the electric field and permittivity ε is a scalar.
I have no idea what they mean by a linear and isotropic medium.. How am I suppose to know if the medium is linear or isotropic?
Nevermind, I figured it out.. But if someone has a better way please let me know.
This is my solution.
Lx + my + pz = 2p
Lx + my + pz = 2L + 2p
Lx + my + pz = 2L + 2m
Lx + my + pz = 2m
L^2 + m^2 + p^2 = 1
Then I looked at the right side of the first four equations and noticed they are all...
The solutions have came up with 5 equations, I'm not confused how they got those 5 equations but I don't understand how it was concluded that L = 0 and m = p = 1/√2.
I am not understanding how Rth is calculated..
What I've been doing is visualizing the path that the current would make. Rth is then the equivalent resistance of those resistors that current travels through.
Is that how it's suppose to be done?
Anyways... if you look at my pictures... the...