An electric field (sometimes E-field) is the physical field that surrounds electrically-charged particles and exerts force on all other charged particles in the field, either attracting or repelling them. It also refers to the physical field for a system of charged particles. Electric fields originate from electric charges, or from time-varying magnetic fields. Electric fields and magnetic fields are both manifestations of the electromagnetic force, one of the four fundamental forces (or interactions) of nature.
Electric fields are important in many areas of physics, and are exploited practically in electrical technology. In atomic physics and chemistry, for instance, the electric field is the attractive force holding the atomic nucleus and electrons together in atoms. It is also the force responsible for chemical bonding between atoms that result in molecules.
Other applications of electric fields include motion detection via electric field proximity sensing and an increasing number of diagnostic and therapeutic medical uses.
The electric field is defined mathematically as a vector field that associates to each point in space the (electrostatic or Coulomb) force per unit of charge exerted on an infinitesimal positive test charge at rest at that point. The derived SI units for the electric field are volts per meter (V/m), exactly equivalent to newtons per coulomb (N/C).
Homework Statement: The amplitude of the oscillating electric field at your cell phone is 4.0 μV/m when you are 10 km east of the broadcast antenna. What is the electric field amplitude when you are 20 km east of the antenna
Homework Equations: electric field
i've done
E=##\frac A...
i've started from this I1=I2
then
I1= JA1=##\frac {E l} R##
I2= JA2=##\frac {E_2 l} R##
but can't get anything useful relating them. Am i forgetting any other useful formula?
I get as result E4
Here is picture. Answers is A.
My attempt was that I thought if i were to place a positive test charge then it would go from top to bottom if there was a positive charge in the center it was avoiding and a positively charged particle at the top, but an electron at the bottom so it would avoid...
Homework Statement: uniformly charged disk, radius r, with surface charge density ##\sigma##
. I want to find the electric field along the axis through the centre of the disk at a h distance
Homework Equations: ##dE=\frac {kdq}{r^2}##
My Solution:
##dE=\frac {kdq}{r^2}##
in this case r=s...
Homework Statement: A thin rod of length L and charge Q is uniformly charged, so it has a linear charge
density ##\lambda =q/l## Find the electric field at point where is an arbitrarily positioned
point.
Homework Equations: ##dE=\frac{Kdq}{r^2}##
A thin rod of length L and charge Q is...
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...
Which is better to use? The equation for the area or the circumference of a circle?
Schaum's Electromagnetics (4 ed) by Edminister
vs
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elecyl.html
I am having trouble solving the following problem. I am given two positive charges on the x axis:
I know that the electric field strength at point P is ##E=150 \frac{V}{m}##, ##d=1.8m## and ##a=2.5m##. I want to find the charge of ##Q##. As far as I know, the electric field on the y-axis...
The contribution coming from a little segment of the ring is ##d\vec{E}=\frac{dQ}{r^2}cos\theta \hat{z}##, assuming that the horizontal components cancel out. But how can we show that?
I believe the answer is incorrect, reasons:
The answer assumes that electric field will exist .
But this is not the case , until and unless there is a bipolarity there cannot be an electric field ( in case of isolated charged objects, the field exists because the bipolarity is separated by a...
When a charge is at rest, it has an electric field only. When the charge starts moving , it is said to have accompanied a magnetic field. My question relates to its electric field while in motion. Does it still exist or not? I know in electron guns electrons are deflected while passing thru the...
We are given: q1 = +2.0 x 10-5 C, q2 = q3 = -3.0 x 10-5 C, r31 = r21 = 2 m
a) We start by finding the electric force between q3 to q1 and q2 to q1
FE31 = k * q1 * q3 / r312
FE31 = (9.0 x 109 Nm2/C2) * (+2.0 x 10-5 C) * (3.0 x 10-5 C) / (2 m)2
FE31 = 1.35 N
FE21 = k * q1 * q2 / r212
Since...
I know how to find the electric field of more traditional designs i.e. a sphere, through Gauss' Law but I don't think Gauss' Law applies to this scenario. I tried to separate each part of the electrode into simple spheres and rods and using Gauss' Law to find these individual elements. This...
If you place a zinc / copper battery it will create an electric field pointing from the copper to the zinc and my question is this what makes this electric field , the zinc pushes electrons in the circuit and never "stays" negatively charged for an electric field to be created same for the...
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...
Options are at the top of page as a) b) c) d)
Answer may more than one.
Now since 'a' is distance from the smaller surface of cone so as we move along the axis area will increase,So current charge density will decrease and as we know J=sigma E,E will decrease,but V will remain constant since...
So I understand that when an electric field is produced in a conductor of length L, the net electric field in the conductor will be 0 because the rearrangement of electrons in the conductor results in the production of its own electric field which cancels out the one produced initially...
Ex =10.0Vm-1
dx= d2=x2+y2
x=4.00m
y=3.00m
d2=4.002+3.002
d=5.00m
Ex = -dV/dx
10.0Vm-1=-dV/5.00m
10.0Vm-1*5.00m=-dV
-50.0V=dV
So from origin at 100V-50.0V = 50.0V
But the solution I am given gives the answer at 50.5V and the information "directed at 45.0◦ " does not seem to have been used so may...
I read Wikipedia's description of how a plasma ball works. Question: What kind of energy is the "radio-frequency energy from the transformer"? Is in the form of electric field energy, magnetic field energy, or both? Thank you!
(from Wikipedia)...
Although many variations exist, a plasma lamp is...
I bring a dielectric in a region with electric field ##\vec E_0##. Net electric field ## \vec E_{net} = \vec E_0 + \vec E_p ## , where ## \vec E_p ## is electric field due to polarization of dielectric.
For linear dielectric, ## \vec E_p ## is 0 outside the dielectric. So,
## \vec E_{net} =...
I tried to find it the following way but to no avail:
Let maximum value of ##\sigma## be ##S##
Now unfortunately, we do not have a maximum value for ##\dfrac{1}{r^2}## because the field point can be as close as we want to the arbitrary surface charge. (The field at a point on the surface is...
Given $$\vec E = -\nabla \phi$$ there $$\vec d \rightarrow 0, \phi(\vec r) = \frac {\vec p \cdot \vec r} {r^3}$$ and ##\vec p## is the dipole moment defined as $$\vec p = q\vec d$$
It's quite trivial to show that ##\nabla \times \vec E = \nabla \times (-\nabla \phi) = 0##. However, I want to...
Hi,
A solution contains some ions (charged particles). We are only interested in my exemple to positive ions.
It is assumed that these ions acquired some mobility under a concentration gradient. Their direction is A to B.
Then these ions encounter/cross an electric field which is oriented from B...
The electric field due to a dipole distribution in volume ##V'## can be viewed as electric field due to a volume charge distribution in ##V'## plus electric field due to a surface charge distribution in boundary of ##V'##.
##\displaystyle\mathbf{E}=\int_{V'} \dfrac{\rho...
Hi!
This is more a conceptual question rather than the calculation itself.
So, Griffiths' section 4.2.1 "The field of a polarized object / Bound charges" says that if you want to calculate the field produced by a polarized material, you can find it from the potential of a surface charge and a...
Hi,
I've a question about electricity in the following scenario: consider an accumulator (e.g. a 9V battery) and an analog/digital voltmeter having a probe connected to the accumulator + clamp and the other to the ground (for instance connecting it to a metal rod stuck in the ground).
Do you...
I know how the answer is C, since E=F/q and F=ma=mg. However, I am a bit confused as to why my other method doesn't work. I thought that since the droplets are falling at a constant velocity, there is not net force, so according to E=F/q the electric field must be zero then? This seems like a...
Two identical conducting spheres A and B carry equal charge. They are separated by a distance much larger than their diameters. A third identical conducting sphere C is uncharged. Sphere C isfirst touched to A, then to B, and finally removed. As a result, the electrostatic force between A and B...
I uploaded a diagram of the problem.
I treated this as many thin rings and integrated it over the length. I placed my origin as in the same place as the uploaded picture.
Finding the electric field due to one small ring:
##\vec r =\langle w-x, 0, 0 \rangle## where ##x## is the distance of the...
Tried to use gauss law but there isn't any usefull symmetry that I have seen. Also tried to integrate the field due to small charges over the whole cube, didn't work too since the integral were too much complicated.
How can a negative charge move towards a position of a higher electric potential from lower potential but lose electric potential energy?From my understanding, I understand that for a positive charge, it must lose potential energy from the electric field as work is done by the electric force in...
Hi. Need help with physics homework. I was able to separate each term and find sigma from the second term and possibly lambda from the first term. Not sure how to approach the third term. (attached attempt at question)
Hi. Need help with physics homework. I was able to separate each term and find sigma from the second term and possibly lambda from the first term. Not sure how to approach the third term. (IMAGE ATTACHED)
I'm looking for a high voltage power supply. I have no experience with such a power supply, nor with all the terms or specifications used for such tools, so I'm looking for general suggestions to what to look for.
I want to generate an electric field or potential field between two points a few...
Homework Statement
Capacitor has a charge of 0.008C. Capacity is 2 μF. Need to calculate energy of electric field. Answer: 32J
Homework Equations
E = q2/2*C
The Attempt at a Solution
This seems right equation, but instead of 32J I get 16J.. any ideas?
1. Homework Statement
the a shell is charged Va=120v
shell b is grounded, Vb=0V
What is the voltage in the center of shells (vo)?
The electric field in the center of shells?
The potential energy in the center of shells?
Homework Equations
Vr=Va+(1/r-1/a)/(1/a-1/b)Vab (from integrals)
though I...
If a charge is put inside a spherical shell, why is the electric field outside the shell independent of the location of the charge? Gauss's law could find that the net flux is independent, but not each individual field?
Is this something about the surface charge density being independent of...
Hi.
The derivation of the capacity of an ideal parallel-plate capacitor is inconsistent: On the one hand, the plates are assumed to be infinitely large to exploit symmetries to compute an expression for the electric field, on the other the area is finite to get a finite expression for the...