The electric potential (also called the electric field potential, potential drop, the electrostatic potential) is the amount of work energy needed to move a unit of electric charge from a reference point to the specific point in an electric field with negligible acceleration of the test charge to avoid producing kinetic energy or radiation by test charge. Typically, the reference point is the Earth or a point at infinity, although any point can be used. More precisely it is the energy per unit charge for a small test charge that does not disturb significantly the field and the charge distribution producing the field under consideration.
In classical electrostatics, the electrostatic field is a vector quantity which is expressed as the gradient of the electrostatic potential, which is a scalar quantity denoted by V or occasionally φ, equal to the electric potential energy of any charged particle at any location (measured in joules) divided by the charge of that particle (measured in coulombs). By dividing out the charge on the particle a quotient is obtained that is a property of the electric field itself. In short, electric potential is the electric potential energy per unit charge.
This value can be calculated in either a static (time-invariant) or a dynamic (varying with time) electric field at a specific time in units of joules per coulomb (J⋅C−1), or volts (V). The electric potential at infinity is assumed to be zero.
In electrodynamics, when time-varying fields are present, the electric field cannot be expressed only in terms of a scalar potential. Instead, the electric field can be expressed in terms of both the scalar electric potential and the magnetic vector potential. The electric potential and the magnetic vector potential together form a four vector, so that the two kinds of potential are mixed under Lorentz transformations.
Practically, electric potential is always a continuous function in space; Otherwise, the spatial derivative of it will yield a field with infinite magnitude, which is practically impossible. Even an idealized point charge has 1 ⁄ r potential, which is continuous everywhere except the origin. The electric field is not continuous across an idealized surface charge, but it is not infinite at any point. Therefore, the electric potential is continuous across an idealized surface charge. An idealized linear charge has ln(r) potential, which is continuous everywhere except on the linear charge.
Hello! I do it for definition because Gauss's law not possible.
I use cartesian coordinates:
$$dq=\lambda dx'\text{because is only in x direction}$$
$$\vec{r'}=(x',0)$$
$$\vec{r}-\vec{r'}=(x-x',y)$$
$$\hat{r}=\frac{(x-x',y)}{\sqrt{(x-x')^{2}+y^{2}}}$$...
Hi, I found this interesting video about How electricity actually works.
The point he makes (see for example the video at minute 7:23) is that the energy in a light bulb connected to a battery is actually transferred by the electromagnetic field and not by electrons flowing through it. The...
I reviewed some of the fundamental physics and I looked back at the equation for Electric potential at a point p:
$$V(p) = k \sum_{i} {\frac {q_i} {r_i}}$$
where
- p is the point at which the potential is evaluated;
- ri is the distance between point p and point i at which there is a nonzero...
Hi! I am a very lost physics student here.
I got a) but I have no idea how.
The formula I used was from an online source and it was:
I think I need a contextual explanation of this formula before I attempt b).
My understanding of electric potential is that it is NOT potential energy, but...
I solved using the formulae listed in the relevant equations and got the right answer.
However, I noticed something strange to me.
The electric potential due to the inner semi-circle was equal to that due to the outer semi-circle.
But based on the formula for calculating V, we notice that there...
We have a parallel plate capacitor with two different dielectrics
It seems to be the case that the potential difference on each half of the capacitor is the same.
Initially, the electric field was ##\vec{E_0}=\frac{2\sigma_+}{\epsilon_0}\hat{j}##.
If we were to insert a single dielectric...
The question says:
Find the electric potential of the infinite line charge at ##\Phi \left(x,y\right)##, when known ##\Phi \left(0,0\right)=0##
I am having hard time finding the electric potential of such.
We know that the line charge is infinite at Z axis.
And we know ##\Phi...
Electric potential = "absolute potential"
Textbooks usually connect both ends of two capacitors, of different voltages, in parallel. What would happen if we only connect one end of the capacitors? Perhaps we would have to solve for Maxwell's coefficients of potential for these two cases (to...
Here is a depiction of the problem
a) The potential at any point P due to a charge q is given by ##\frac{kq}{r}=\frac{kq}{\lvert \vec{r}_s-\vec{r}_P \rvert}##, where ##r## is the distance from the charge to point P, which is the length of the vector difference between ##\vec{r}_s##, the...
I only could calculate the distance travelled by each body, by making the difference between the initial and final electric potential work equal to the work of friction done by the 2 bodies.
I tried to make the kinetic energy of the first electron equal to the electric potential work.
mv^2/2=ke^2/d
We have to solve for the minimum distance between them: d=2ke^2/mv^2=5.05*10^-10 m
The force is: F=ke^2/d^2=9*10^-10 N, which is not correct.
Can anyone explain to me why grounded means zero electric potential. I confuse what's the relation between infinite ground conducting plane and its electric potential (the method of images).
I have a several question:
1. Why the conductor plane must be infinite, while in reality there's no...
For A.1 of this problem,
The solution is
However, I have a doubt about the linear charge density ##\lambda##.
I don't understand how ##\lambda = \frac {q}{2\pi R} ## since this is not a thin ring, but has a non-negligible width of ##2a##
I think that the toroid has a larger area than thin...
For part (a) of this problem,
The solution is
However, my solution is
Am I correct? In the solutions that don't appear to plot the electric potential as units of ## \frac {k_eQ} {a} ## like I have which the problem statement said to do.
Many thanks!
Why do we have a charge in the denominator of equations for voltage and el. potential if both voltage and el. potential are not dependent on charge?
Is it just because that was the only way to derive the formula for voltage and then we realized we don't need q? U=W/q --> U=eqd/q.
I can calculate the electric field strength at any point above the plane with Gauss' Law (##E = \frac{\eta}{\varepsilon_0}##) and so the electric potential at any point a perpendicular distance ##z## above the conducting plane (##V=−\frac{\eta}{\varepsilon_0}z##).
But I'm having trouble taking...
I've already tried to calculate the potential with respect to the 3 segments and then apply superposition (V1+V2+V3). However, I was not very successful. My error I think is in the calculation of the radii, mainly of the line segment that is on the z axis. Can anybody help me? I need some light...
We can find the potential energy by finding the potential difference between the two masses. the minimum distance between the two masses is 10 cm. The maximum is 30 cm because they can be 3 string lengths apart as they repulse each other once the string is cut.
So, to get potential difference...
I am able to get V1 = kq/a - 4kq/b
and V2 = kq/b + -4kq/b
For some reason the solution says it is V1-V2 as opposed to V2-V1.
Maybe has something to do with positive shell in the center and negative outer shell? I know the electric field goes from positive to negative, but I don't know how...
By measuring angle \theta from the positive ##x## axis counterclockwise as usual, I get ##d\vec{E}=k( (\lambda_2-\lambda_1)\cos(\theta)d\theta, (\lambda_2-\lambda_1)\sin(\theta)d\theta )## and by integrating from ##\theta=0## to ##\theta=\frac{\pi}{2}## I get...
According to theory I should be able to get the Electric Field (E) from its pOtential (V) by doing the grad (V) so
E = -grad(V), however, V is contant V = k*lambda* pi which results having E =0, but this is not right. What I am missing??
see figure below
The answer should be Ex = 2*k*lambda / r...
(a) Knowing ##E##, we can use equation (2) to determine ##V##. However, since ##\vec E## represents the distribution of electric field in space i.e. a function of (x,y,z). For example, ##\vec E = x \hat i + y \hat j + z \hat k##. Here we do not know this function so how can we know ##V## at a...
u = (9*10^9)(1.61*10^-19)^2 * (1/[3*10^-15 ]- 1/[2*10^-10])
u = 7.68*10^-14 J
but here the question. I have been taught that W= -U so shouldn't the answer be negative??
When i look up at the solution all other sources say that the W = U and therefore the answer is in postive.
I have been having a hard time understanding Electric Potential and believe I finally have a grasp on what is trying to say. I wanted to right out my understanding here and hopefully have someone confirm what I am saying is somewhat accurate as I feel like when you write stuff out you tend to...
I set up an equation for the sum of all the potential energies and when cancelling out ##k## and ##q^2##, I got ##\frac{1}{0.05}-\frac{1}{x}-\frac{1}{0.05-x}=0##. However, this has no solutions, so I must've gone wrong somewhere. Could someone just give me a hint, not a solution, that would put...
I was wondering, we constantly assume the reference of zero potential is the surface of the Earth. But if we consider the reference to be the infinity, what would be the electric potential of the Earth?
As Faraday says, the Earth is charged with a -580 kC of negative charge. If we consider...
So I have a ring(red) of uniform charge ##\lambda## per unit length, and I want to calculate the electric potential at the origin (actually on any point of the ring). It is clear that the ring is given by the equation $$r=2 R \sin \theta$$, in polar coordinates, where R is the radius of the...
I am trying to work out the co-rotating electric potential ##\Phi = \xi^{\mu} A_{\mu}## for the KN solution. First it's necessary to prove that the hypersurfaces ##r = r_{\pm}## are Killing horizons ##\mathcal{N}_{\pm}## of a Killing field of the form ##\xi = k + \Omega_H m## for some Killing...
I tried to find the charge distribution using the given potential but couldn't produce the correct result. Also, Gauss's Law doesn't help, as the electric flux is 0 but we don't have any symmetry. Can someone please shine a light on this? Thanks in advance..
So this is more of an intuitive question rather than a mathematical one. I present the problem.
Assume I have 2 charges of charge +q at a distance r from each other on the z axis. Position of two charges is (0,0,r/2) and (0,0,-r/2). Assume now that I want to calculate the force these two...
In my book, the potential gradient for a charge placed anywhere in space is defined as: E = -V/r
HOWEVER, for parallel plate (capacitors) the potential gradient is defined as E = V/d (V being the potential difference). How come there's no negative sign for the potential gradient of the parallel...
1- Write down the complete MAXWELL equations in differential form and the material equations.
2- An infinitely extensive area is homogeneously filled with a material with a location-dependent permittivity. There are charges in the area. Give the Maxwell equations and material equations of...
So, each capacitor must have a different potential difference, given by its capacity and charge... this would cause charge and current accordingly to flow in the circuit.
But how do I determine the final potential difference, which would of course be the same for both of them? I have tried...
So, having two parallel resistor ##R_{1}## and ##R_{2}## , the current flowing through the equivalent one will be ##I_{eq}=I_{1}+I_{2}##.
Now, it comes the point I'm not totally getting: why is ##V_{eq}=V_{1}=V_{2}##? These V's are the difference of potential measured between which points...
Specifically, I haven't really got all the "methods" through which you could calculate or derive the electric potential and in some situations, I cannot understand how and when to apply this concept.
Is it something caused by any charge, or must there be an interaction between the two to...
My first attempt revolved mostly around the solution method shown in this "site" or PowerPoint: http://physics.gmu.edu/~joe/PHYS685/Topic4.pdf .
However, after studying the content and writing down my answer for the monopole moment as equal to ##\sqrt{\frac{1}{4 \pi}} \rho##, I found out the...
Ve=0m/s
Vp= 0m/s
Qe/Qp= 1.60E-19
Me=9.11E-31
Mp-1.67E-27
Ive pretty much gathered all of the equations I think I need to solve the problem. I just am stuck. The last step I realize that the forces would be equal to each other so I have mp x ap = me x ae but then when I try to solve for the...
I used the potential at the surface of the sphere for my reference point for computing the potential at a point r < R in the sphere. The potential at the surface of the sphere is ## V(R) = k \frac {Q} {R} ##.
To find the potential inside the sphere, I used the Electric field inside of an...
The problem is for a solid sphere uniformly charged with Q and radii R.
First I calculated taked ##V(\infty)=0##, giving me for :
$$
\begin{align*}
V(r)=&\frac{3Q}{8\pi\varepsilon_0 R}-\frac{Q}{8\pi\varepsilon_0 R^3}r^2\qquad\text{if $r<R$}\\
V(r)=&\frac{Q}{4\pi\varepsilon_0 r}\quad\text{if...
Let us connect a battery of potential difference V to a wire. There is no resistance. Nothing!
Now the battery creates some potential difference and the charges in the conducting wire move due to the Electric field created in the conductor by the battery. So, as the charge moves, its potential...
Hi,
I think this problem is solved in exactly as a similar problem where the two spheres are very far apart and connected by a very long thin conducting wire. I'm trying to explain this in words, since LaTeX does not seem to work any more (for some reason LaTeX syntax is not replaced by maths in...
Not sure how the problem set up initially as no diagram was provided in the question. Please help me to start with the solution. Your assumptions and educated guess are appreciated.