# Tag Info

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We know that field due to point charge is zero at a distance infinity (and beyond) from it. Let us consider a square with equal charges at the corners. Let us assume that the length of the side of the square to be such that, electric field due to charges at corners of that side, becomes zero (as like field is zero at infinity and beyond it) at some ...

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In addition to Ali's answer, here are some pictures which may be helpful in convincing people that the origin is not the only point inside the polygon where $\mathbf{E}=\mathbf{0}$. Letting the charges be located at $(\cos(2\pi k/N),\sin(2\pi k/N))$ for $k\in\{1,2,...,N\}$, we can generate plots of $|\mathbf{E}|^{-1}$ for various $N$. The zeros of ...

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One can do the calculation(expand the potential to the second order around the center) and show that the center of the polygon is a minimum of potential. We are free to choose $V(\infty)=0$, if we do so, then it would be easy to show that the potential at the center of the polygon is positive. Combining the results above with the fact that the potential is ...

1

Yes, Coulomb’s law is only for point charges separated by a distance $r$. The inverse square law is there because of the diverging or converging nature of electric field from a point charge which is the crucial point. Here is how I explain: Imagine spherical surfaces with increasing radius around a point charge. As we increase the radius, the intensity of ...

1

Gauss theorem is of great importance. Those situations, in which the calculation of electric field by applying Coulomb's law or the principle of superposition of electric fields becomes very difficult, the results can be obtained by applying Gauss's theorem with great ease. You can notice that electric field due to a point charge decreases inversely as ...

0

As an idealization if plates are assumed to be infinite in extent there is no field between A and C since they are both positively charged and connected. So q1=q2=0, and q3=q4=3Q/2. Using your equations we get the same result. Grounding means q1+q2=0, so the same conclusion.

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Coulomb's law is indeed a special case between two point charges. to find the force between a point charge and a plate, you would have to integrate the equation over the plate surface to calculate the contributions from all infinitesimal charge elements. It's more practical to figure out what the electric field is, and then use $\vec{F}_e = q\vec{E}$ to ...

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You start with a total of $3Q$ on the outer plates - and that charge has nowhere to go. So when you connect the two plates together, they will share the charge equally, and have a net charge of $1.5Q$ on each. Grounding the center plate means that charge can flow freely. So what will happen? First - charge will flow between plate B and ground. The boundary ...

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You are thinking about energy, and not about force. The charge exerts a force upon the system, or the system exerts a force upon the charge. Energy is consumed in that process, but that isn't the point of the statement made. Look at it through Energy $E$.(the field will be $\mathbf E$) Lets look at these formula: \begin{align} E&=\int_\ell \mathbf F ...

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Your teacher is right. Although you missed a factor of $2$ in last line. The case you are saying is NOT equivalent to this one. How do you plan on filling dielectric $K_1$ on the lower half which is already filled with $K_2$? This makes no sense.

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To move charge from one point to another in an electric field, the force which we must apply is equal and opposite to the force due to the field. The sentence you provided is actually confusing. I think it should have been: To move charge from one point (lower potential) to another (higher potential) in an electric field, the constant force which ...

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Remember that, if the net force is zero, velocity is constant (not necessarily zero!). You only need to do push a tiny bit harder for a tiny bit of time to start moving the charge. This extra amount can safely be ignored. Once the charge is moving with some nonzero velocity, equal force is enough.

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Plasma is a state of matter which is composed entirely of charged particles or ions. By this definition, a flame is a plasma in itself because it consists of charged particles which can be demonstrated by placing a flame in a uniform electric field. The flame bends to a side in presence of a field. It may also be due to the fact that an electric field ...

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Coulombs date back to the 1860's, and even predate CGS units. The connection between the volt-ohm-second and mks units were made only in 1904. Coulomb used e.s.u. based on the french ft lb s system. The coulomb constant is a feature of choice of units, if charge is found from LMT, then the size of the coulomb constant can be set to one.

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No, Coulomb did not know the Coulomb as a unit. According to this page, the Coulomb was defined at the 9th CGPM (General Conference on Weights and Measures) conference, in 1948. Wikipedia gives the same date. Coulomb could not measure charges, but he could create a charge and then halve it, quarter it, etc by letting the charge flow from one object to ...

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To address the follow up question, which requires more characters than a comment allows: In a simple idealized view, the Fermi level is the top energy level in the solid occupied by electrons. In silicon with no doping it sits at mid-gap: the valance band is full, the conduction band empty. In a thought experiment, if you had two separate chunks of ...

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Since plate B is grounded, charged conductors near it can induce equal and opposite charge on plate B. Infact a ground is necessary to charge an object. http://www.physicsclassroom.com/class/estatics/Lesson-2/Charging-by-Induction When a charged conductor is brought near a grounded conductor opposite charges are induced on the grounded conductor but overall ...

2

I presume you are referring to process of nuclear fission of uranium-235, which has the equation: $$^1_0\text{n}+^{235}_{\ \ 92}\text{U}\longrightarrow ^{236}_{\ \ 92}\text{U}$$ However, a subsequent reaction is: $$^{236}_{\ \ 92}\text{U}\longrightarrow^{144}_{\ \ 56}\text{Ba}+^{89}_{36}\text{Kr}+3^{1}_{0}\text{n}$$ The production of neutrons is a feature of ...

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You didn't really simplify the expressions in the same manner, so they're hard to compare. Even if they're not the same, it's often useful to have a clear idea of what differs in the final result. Simplifying the first expression to share the denominator $(b^2-c^2)^2$: \begin{align} W&=\frac{1}{2}\int\rho V\,d\tau\\ &=\frac{\pi ... 3 The analogy follows with the right definitions. The "flux" of the "vector" E(z) through a contour \Gamma is \mathrm{Re}\left(\int_\Gamma E(z)^*\,\mathrm{d} z\right). I think you may have forgotten the conjugate in the relationship between the "Electric field" and the complex potential \Omega: E(z) = -(\mathrm{d}_z \Omega(z))^*. So it is the ... 4 There is indeed a connection. The holomorphy is easily seen in the electrostatic potential. In a charge free (two-dimensional) region, the electrostatic potential solves Laplace's equation and hence is a harmonic function. The real and imaginary parts of a holomorphic function are harmonic functions and thus the electrostatic potential can be identified ... 4 You can use q/\epsilon_0 to calculate flux for both cases because that's what Gauss' law says: Just look at the enclosed charge. It's amazing. Flux for any closed surface is \Phi = \oint \vec{E} \cdot d\vec{A}. There are two ways to calculate this quantity: The hard way, which means evaluating \vec{E} on every part of the surface, and integrating. ... 1 But why when I connect a volt-meter across the whole diode will I not see this built in potential? Because at equilibrium (V=0) the Fermi-level throughout the whole device is flat. A voltage will only exist when there is a gradient in Fermi-level over the component (i.e. a voltage drop). Edit Maybe a more visual explanation would help? Think of the ... -2 Actually, B wouldn't stay at 0V since the other plates would induce a charge on B. Since you have q_1 = 2q_2, you should be able to figure out the charge distribution on A and C. 0 The electric potential at a point is actually defined as such: V={PE \over q} $$Therefore:$$PE=qV It is not derived from the equation of the force between two point charges. Since you are familiar with the equations for the relationships between point charges, however, you can conceptualize the above equation like this: Usually I think of the ... 0 PE = qV is usually used in parallel plates. V = \frac{kQ}{r} is the electric potential at a point, r distance from the charge Q. So given the charge and the distance from a point, you use the second formula. V = \frac{kQ}{r} was derived from PE = \frac{kqQ}{r} and PE = qV. I hope you get the point I'm trying to make 0 The line integral would only work for finding a potential difference, which I presume is what you need. Since the line integral needs a path(to perform the integration), hence, it would need a general expression for electric field(in your case, finding the potential requires integrating from \infty to A, so you'd need the expression for all points ... 1 The whole (pedagogical) point of the slide wire generator is to illustrate that not only do changes in the magnetic field generate current in the loop, changes in the area of the loop - in a constant magnetic field - also generate a current. It's the change in magnetic flux that matters. As long as the wire is moving with some velocity, the magnetic flux ... 1 No, your reasoning is incorrect, because there's no reason for the forces to cancel in general. Actually, the charge in general will be attracted by the field of the induced opposite charge on the inside surface of the conductor. This is easy to see by use of the fact that \nabla^2 V=0 in the region devoid of charges implies that V is a harmonic ... 3 The gravitational field \mathbf g equals, by definition, negative of the gradient of a correspondonding potential \Phi; \begin{align} \mathbf g = -\nabla\Phi. \end{align} Therefore, it suffices to produce a gravitational potential \Phi whose value is zero at a point but whose gradient is non-zero at that point. This is straightforward to do. Let a ... 1 It does make sense to talk about the number of field lines, but only if you take care to represent the field amplitude as being inversely proportional to the spacing between the lines. With that, the total number of lines crossing a surface is proportional to the flux, etc. Some people, notably textbook authors Chabay and Sherwood, feel that the field line ... 0 Since the force is directed from q_0 to q_1, you should also consider the case when d_1 is negative. A combined answer could be written as: \vec F=\frac {kq_0 q_1}{d_1^2}(d_1/|d_1|) \hat{j} 1 That is correct. Coulumb's Law can be written F = k \frac{Q_1 Q_2} {r^2} \hat{j} $$where your  d_1^2  is the distance labeled as  r^2 , and  k = \frac {1} {4\pi \epsilon_0} .  \epsilon_0  is the permittivity of free space:  8.85418782 × 10^{-12} m^{-3} kg^{-1} s^4 A^2  0 The force on the charge q is given by the electric field \mathbf E_m = -\nabla \phi_m of charges of the metal shell surrounding it. This field does not vanish inside the metal, because total field \mathbf E_m + \mathbf E_q does. It follows that the potential of metal charges \phi_m is not constant throughout the metal and its inner surface is not ... 1 The answer to your question is "no". Put the point charge close to the wall at some spot of the wall. There will a surface charge be collected at this spot of the wall that attracts the point charge. In the following I give an example for which one could even calculate the attracting force analytically. Nevertheless, I keep a bit informal here since the ... 0 I think U must be the potential energy to bring the last particle in from infinity. Thus U/5 is the average pairwise potential between this last bead and each of its neighbors. By symmetry though, each bead has this same average pairwise potential with the other beads. The total potential of the configuration is half the sum of the pairwise potential of ... 3 The switch really has 2 positions: on and off. However, when you move the switch very slowly, it may leave the closed position slowly. When the switch is just barely open, the field may cause the air to break down and start conducting, to form a spark (as @anna v explained). To rephrase, the reason why sparks happen is because the switch may only be open a ... 0 Field lines are tangential lines obeying the Gaussian law. The can only meet at an angle at discontinuities in the field like point charges or surfaces with inhomogenous magnetic flux and similar. 2 Air is a bad conductor up to a certain value of the field generated by charges and the distance between them. After that air breaks down and a discharge happens, i.e. sparks. So below this level charges can accumulate by rubbing for example , positive ions left on one surface and negative on the other. When brought close a spark occurs. Why does holding ... 2 Field lines are a visual representation of a mathematical construct, like a graph of a function. The defining properties of this visual representation are The field lines run parallel to the field at every point. The density of the field lines in an area is proportional to the strength of the field. The second property tells you that the field lines can ... 4 Electric field lines reveal information about the direction (and the strength) of an electric field within a region of space. If the lines cross each other at a given location, then there must be two distinctly different values of electric field with their own individual direction at that given location. This could never be the case. Every single location in ... 13 Electric field lines are a visualization of the electrical vector field. At each point, the direction (tangent) of the field line is in the direction of the electric field. At each point in space (in the absence of any charge), the electric field has a single direction, whereas crossing field lines would somehow indicate the electric field pointing in two ... 5 The electric field at any point is the sum of all the fields due to each individual charge in the system. The field has a magnitude and a direction. The field lines are a representation of the magnitude and direction of the field over an illustrated area. The field lines point in the direction of the field. If lines from two sources were to cross, we could ... 0 Should this have a homework tag? If the radius of the cable is r_0, the charge density \rho = \alpha\delta(r -r_0). You need a constant of proportionality that makes the total charge per unit length be \lambda. The total charge in length l would be q = \lambda l = \iiint \rho dx, where the integral is over length l along the z axis. If you ... 0 Seeing your comment, it seems you are concerned about group of charges with certain mass. Then you need to apply Gauss law for the cases where it becomes difficult to apply coulombs law or principle of superposition. In case of gravitational force, find the center of masses of either configuration and you can proceed to find force using Newtons law of ... 0 The magnitude of gravitational force between them would be:$$G\frac{mm'}{d^2}$$The magnitude of electrostatic force, if the magnitude of charge on them is q and q' respectively, would be:$$\frac{1}{4\pi \epsilon _o}\frac{qq'}{d^2}$$The net force would just be the vector sum of the two. 0 Since k_e is of the order 10^{9} and G is of the order 10^{-11} thus gravitational force can be neglected, even if you add then can be added safely thus F_T = F_g + F_e = G\times\frac{m_1\times m_2}{d^2} + k_e\times\frac{q_1\times q_2}{d^{2}} 6 Charge is a quantity which arises from Noether's theorem, due to continuuous global symmetries (up to a total derivative) of an action, and as such we have many types of charge, other than electric. For example, consider the Dirac Lagrangian,$$\mathcal{L} = \bar{\psi}(i\gamma^{\mu}\partial_{\mu}-m)\psi which describes fermions. It is invariant by a ...

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Charge is a fundamental conserved property of particles. It is, if you like, a measure of how much a particle interacts with electromagnetic fields. A particle with charge can produce and be affected by electromagnetic fields. This is what we mean when we say a particle has charge. Its a simple quantised way to measure the coupling strength of particles with ...

0

I thought static electricity doesn't really build up when there's a lot of moisture, so that's why you don't get static shocks in the summer when there's more humidity. Even though it's winter now, I figured I wouldn't feel any static shocks if my hands are wet, but I still do. When the air contains more moisture, electron build-up on your body and ...

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