For the first part, since
$$ E(r) \propto \frac{1}{r} \hat{r}$$
by the principle of superposition the maximal electric field should be halfway in between the two wires.
Then I'm not sure how to go about the second part of the question. I understand that the total potential due to the two wires...
F = qE
ma = (2*10^-6) * (λ / (2pi*r*ε0) )
ma = (2*10^-6) * (4*10^-6 / (2pi*4*ε0) ) => Im not certain what to put for r ( But I sub in 4 because dist is 4)
a = ( (2*10^-6) * (4*10^-6 / (2pi*4*ε0) ) )/ 0.1
a = 0.35950
v^2 = U^2 + 2 a s
v = 0
u^2 = -2 a s => Can't sqrt negative so i...
Attached is the subsection of the book I am referring to. The previous section states that the electric field magnitude at any point set up by a charged nonconducting infinite sheet (with uniform charge distribution) is ##E = \frac{\sigma}{2\epsilon_0}##.
Then we move onto the attached...
Here I am going to include the proof provided by my book. It is quite a splendid explanation, though there are a few key points I have yet to fully understand. If the electric force by the electric field on the charge at the surface of the conductor is conservative (which it is), then why is...
Let's say I place a positive point charge inside a hollow conducting sphere. If we take a Gaussian surface through the material of the conductor, we know the field inside the material of the conductor is 0, which implies that there is a -ve charge on the inner wall to make the net enclosed...
So for the Gaussian theorem we know that $$ \frac{Q}{e} = \vec E \cdot \vec S $$ Q's value is known so we don't need to express it as $$Q=(4/3)\pi*(R_2 ^3-R_1 ^3)*d$$ where d is the density of the charge in the volume. I've expressed the surface $$S=4\pi*x^2$$ where x is the distance of a point...
The charges are q1,q2 & q. P,Q,O1,O2 refer to positions only. This is a conducting sphere with cavities containing charges.
I'm interested in knowing how the charge should be distributed in the sphere. I know the charges induced on the charges of the sphere should be equal and opposite to the...
I try to do my assignment which is based on mass spectrometer entirely. The mass spectrometer i am working on has these parts below:
1.Accelerator region
2.Velocity selector region
3.Spectrometer
The elements i am working on are isotopes of the same element and they all enter the accelerator...
Electric currents and the things within are generally explained through the help of intuitive water current examples, where potential difference is explained through the pressure difference and electric current is explained as the flow of water. But I like to think in terms of some driving force...
I think the right solution is c). I'll pass on my reasoning to you:
R=6\, \textrm{cm}=0'06\, \textrm{m}
\sigma =\dfrac{10}{\pi} \, \textrm{nC/m}^2=\dfrac{1\cdot 10^{-8}}{\pi}\, \textrm{C/m}^2
P=0'03\, \textrm{m}
P'=10\, \textrm{cm}=0,1\, \textrm{m}
Point P:
\left.
\phi =\oint E\cdot...
I thought the right choice was d). But when it comes to the solutions, it is b) and I don't understand why.
My reasoning would be: the potential at a point is the work that the electric field does to transport a charge from infinity to that point, so if the field is zero, it does no work and...
Homework statement:
Find the electric field a distance z from the center of a spherical shell of radius R that carries a uniform charge density σ.
Relevant Equations: Gauss' Law
$$\vec{E}=k\int\frac{\sigma}{r^2}\hat{r}da$$
My Attempt:
By using the spherical symmetry, it is fairly obvious...
Please help me to find the position P and whether it will work in this solution by knowing the position.
(The question, my solution and thought in the image)
This page claims that "[t]he electric field outside the sphere is given by: ##{E} = {{kQ} \over {r^2}}##, just like a point charge". I would like to know the reason we should treat the sphere as a point charge, even if the charges are uniformly distributed throughout the surface of the...
The answer according to the key is C. I thought the answer would be E since the electric field inside a conductor is always zero. Can someone explain why the answer is C?
Note that the solution is 5625 V/m in z direction which is found easier using Gauss' law, but I want to find the same result using Coulombs law for confirmation.
Lets give the radius 0.04 the variable a = 0.04m.
##\rho## is the charge distribution distributed evenly on the surface of the...
The first time I saw this question I had no idea how to do it (as you can see in the figure, I lost a lot of points :s) because I was confused on how to even approach it with area of the slab from all sides being infinity. Right? That's problematic, no?
Today, I just tried the problem again for...
Hi! I need help with this problem. I tried to solve it by saying that it would be the same as the field of a the spherical shell alone plus the field of a point charge -q at A or B. For the field of the spherical shell I got ##E_1=\frac{q}{a\pi\epsilon_0 R^2}=\frac{\sigma}{\epsilon_0}## and for...
V(ρ) = V_o*ln(ρ/0.0018)/ln(45/180)
(Attached picture is where the unit vector of r is really ρ.)
In cylindrical coordinates
∇V = ρ*dV/dρ + 0 + 0
∇V =derivative[V_o*ln(ρ/0.0018)/1.386]dρ
∇V = V_o*0.0018/(1.386*ρ)
E = V_o*0.0012987/ρ
Work = 0.5∫∫∫εE•E dv
Bounds: 0.0018 to 0.00045 m
D = εE =...
Hi! My main problem is that I don't understand what the problem is telling me. What does it mean that the surface is a flast disc bounded by the circle? Is the Gauss surface the disc? Does that mean that inside the circle in the figure, there is a disc?
Can you give me some guidance on how to...
So I figured out the potential is: dV = (1/(4*Pi*Epsilon_0))*[λ dl/sqrt(z^2+a^2)]
.
From that expression: We can figure out that since its half a ring we have to integrate from 0 to pi*a, so we would get:
V = (1/(4*Pi*Epsilon_0))*[λ {pi*a]/sqrt(z^2+a^2)]
In that expression: a = sqrt(x^2+y^2)...
I tried to work out both a) and b), but I am not sure if I am correct. I drew a picture with a sphere around q first with radius r and then with radius 3r.
For a) ##E.A=\frac {q}{ε_°}## (when using Gauss' Law)
Since ##A=4πr^2##, I substituted this in the equation and solved for E giving me...
Hi, so I was able to solve this problem by just equating the forces (Tension, mg, and EQ).
But I thought I could also solve this problem with Conservation of Energy.
However, I calculated it several times, and I never get the right answer this way.
Doesn't the Electric Field do the work to put...