# Tag Info

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Without even doing any circuit calculations, you can conclude the voltage between a and b is zero by symmetry. Proof: Assume there's a voltage between the two points. If you close the switch, a current would flow. If you take the mirror image of the circuit, you'd expect the same current to flow, but in the opposite direction. Except the circuit is left ...

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First, let's assume the left-most terminal is connected to the positive terminal of the battery and the capacitor voltage reference direction is left-most terminal positive. Now, consider a KVL loop clockwise through the top 2C capacitor, the switch, the bottom C capacitor and the battery: $$10 \mathrm V = V_{2C_{top}} + V_{ab} + V_{C_{bot}}$$ So, the ...

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The potential difference across the top two capacitors must be the same as the difference across the bottom two. I will number the capacitors $C_{11}$ for top left, $C_{12}$ for top right etc. If we assume the charge on each capacitor is the same, then the voltage difference must be zero. But if we can assume that each capacitor may have a different charge ...

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The one with the greater magnitude or absolute value is greater. The sign has no bearing. Negative work only tells us about the direction in which the work is being done, positive along the direction of motion, and negative anti-parallel to the direction of motion. An object in a gravitational field can escape to infinity if TE>0. GKE has to be greater ...

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To see that your integral expression does not make any sense, imagine that $\vec{r}(t)=( x(t),y(t))$ describes a circle. Then the line integral of the force around the loop gives the change in potential energy, which should of course be zero, $$\oint \vec{\nabla} \phi \cdot \vec{dl} = \Delta \phi =0.$$ But if you insert the actual values from your ...

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You should also match the derivatives at $x=0$ so that they took into account the $\delta$-function. If you take smooth well $V_\epsilon(x)$ and consider small region near zero $(-\epsilon,\epsilon)$ where the "meat" of the well is concentrated you may then integrate the Schrodinger equation at that region, ...

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TLDR: The external electric field through the dielectric causes small displacements on the bound charges of the dielectric. These accumulate as an induced surface charge on the dielectric/vacuum interface. The total electric potential is then given as a sum of the external potential, the induced potential of the induced point-like dipole, and the induced ...

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Let me try to offer a different way to see it based on linearity. Note that your two links are reversed in the situation they consider; your second link has the dipole embedded in the dielectric embedded in vacuum, whereas your first has the dipole in a vacuum embedded in a dielectric. So, equation (3) contains an $\epsilon$ where it should have ...

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There are many conditions for identifying that a vector field $\vec v$ is conservative or not: $\nabla \times \vec v =0$ A conservative field vector is essentially irrotational. $\oint_c \vec v \cdot d\vec r =0$ Work done by a conservative vector field about any closed path $C$ is $0$. $\vec v=\nabla \phi$ A conservative vector field can always ...

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Yes. 1. option: Another (equivalent) criterion: If the work done on a arbitrarily chosen closed path is zero, then the field is conservative. I.e: $$\oint \vec{F} \cdot \mathrm{d}\vec{r} = 0$$ means $\vec{F}$ is conservative. 2. option: More general definition of curl $$\left(\mathrm{curl} \ \vec{F} \right)\cdot \vec{n} = \lim_{A \rightarrow 0} ... 0 So I am not aware whether you have regarded the previous answer properly or you know about this, but there is a uniqueness theorem for the solution of the Poisson equation. Thus this solution is already correct because you have solved the boundary problem: You have found a solution for the Laplace equation in the area where no free charges exist ... 6 The issue is that the formula that connects force and potential gets an extra term when the force depends on velocity {\bf v}. The formula reads (see e.g. Ref. 1)$$\tag{1} {\bf F}~=~\frac{d}{dt} \frac{\partial U}{\partial {\bf v}} - \frac{\partial U}{\partial {\bf r}},$$rather than just$$\tag{2} {\bf F}~=~ - \frac{\partial U}{\partial {\bf r}}. ...

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In general, no justification needs to be made - we can assume whatever we want. If a solution which satisfies both the equations and the boundary conditions is found, it is the only solution. By focusing on a subset of all possible solutions, we can make solving the problem much easier. Now, as we have "guessed" the form of solution (so that the equations ...

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Pulsar has already given a correct answer. In this answer we will use a slightly different (but equivalent) method, and we will keep a watchful eye on possible distributional contributions. George Green uses Gaussian units where Coulomb's constant $k_e=1$. He is considering a ball with radius $a$ and uniform charge density $\rho_0$, i.e. of total charge ...

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So, I think I got it thanks to the tips you two gave me. As you mentioned, my single electric fields weren't incorrect, but I have to put them into the same coordinate system. The electric field for the wire running along the x-axis should look something like this then: ...

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You are right. Voltage is an electric potential difference. The concept of potentials is more general (e.g. gravitational potential) in physics.

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When dealing with infinitely long line charges (basically a cylindrical geometry) calculating the potential relative to infinity becomes a problem. You have to establish a reference (ground/earth) at a finite location. So, your result of an infinite potential difference is not incorrect, although it is confusing the first time you see it. This site ...

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$\vec{k} = k_x \hat{x} + k_y \hat{y} + k_z \hat{z}$, and $\vec{r} = x \hat{x} + y \hat{y} + z \hat{z}$; the coordinate dependence is encoded in the $\vec{r}$. These expressions are in Cartesian components, but if you ever need to calculate this in curvlinear coordinates, the logic would be the same. If $\phi$ is not specified, you can probably assume ...

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What is gravitational potential? Usually a potential is defined as the potential energy per mass or per charge or similar. This is most often seen in relation to electricity or chemistry and less often to gravity. $GPE$ is gravitational potential energy. $GP$ is gravitational potential energy per mass: $$GP =\frac{GPE}{m}$$ Is it defined for the system ...

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Let's first derive the value of $V$ inside the small sphere: $$V_\text{sphe} = \rho\int\frac{\text{d}x'\text{d}y'\text{d}z'}{r'},$$ Where the sphere is sufficiently small such that $\rho$ can be considered constant. We can orientate the axes such that $p$ lies on the $z'$ axis. In spherical coordinates, the integral then has the form \begin{align} ... 0 There's a problem in this equation: d\vec{l}=dy \cdot -\vec{j} $$here dy needs a minus sign. It is easier to see this writing the path. The path of integration is parameterized by:$$ \vec{l}=\vec{P}+(\vec{N}-\vec{P})\frac{y-y_P}{y_N-y_P} = \vec{P}+\hat{j}(y-7a) \\ y_p \le y \le y_N $$Therefore$$ d\vec{l}=\hat{j}dy $$and$$ \int_P^N E_y \cdot ...

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Your copy of Verma has already defined gravitational potential energy previously in (11.3) The gravitational potential energy of a two particle system is $$U(r) = -\frac{Gm_1m_2}{r} \tag{11.3}$$ where $m_1$ and $m_2$ are the masses of the particles, $r$ is the separation between the particles and the potential energy is chosen to be zero when the ...

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Your teacher is wrong. The gravitational potential $V(x)$ is generally defined as potential energy per unit mass i.e. $V(x) \equiv \dfrac{U(x)}{m}$. So for the points where $U(x)$ is zero, $V(x)$ is zero and vice-versa by definition. EDIT: After you added the comment and a snapshot of the book, I realized your book has defined Gravitational Potential in a ...

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Firstly there are tests that can show whether a force field is a gradient of a unknown scalar field. If it isn't the gradient of a scalar field then it definitely isn't proportional to the gradient of that particular scalar field. The test alluded to above is whether the circulation around every closed loop is zero. So if you can find the work done around a ...

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In classical mechanics, potential energy is only defined up to an arbitrary constant, and therefore total energy is only defined up to that arbitrary constant: in addition, the kinetic energy is reference-frame-dependent, and in the case of, say, $1/r^2$ force laws, it may not have a well-defined minimum. For these reasons, "zero energy state" has no ...

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Yes there is! First, electric potential is measured in volts (V) and electric potential energy is measured in joules (J). Now if it sounds familiar is that both tell you about an energy quantity (V=J/C). Indeed, in electromagnetism, the potential is seen as the electric field, multiplied by the distance between the source (for example a point charge) and the ...

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Work is positive when the projection of the force vector onto the displacement vector points in the same direction as the displacement vector(you can understand negative work in a similar way). Let's call the charge that you are trying to move Q. Observe that if you want to calculate the work done by the electric field on this charge, you simply invoke ...

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Its because of the fact that we are touching the Earth, so we have the same potential as the Earth. Without a human in the vicinity of a "patch" on Earth, all we have are equipotential lines of 226V per meter by your estimation. When a human arrives at that place, because of the fact that the Earth tends to make us have the same potential with it(it can ...

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