# Tag Info

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If you treat the 1s ground state's probability distribution as a classical charge density distribution (not really accurate, but I think the simplest way to interpret the problem), then there isn't one. This state is spherically symmetric, so the electric field is always radial and depends only on the radial coordinate and not on the angular coordinates. So ...

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I can't understand your question thoroughly. But if you connect the negative terminal of the battery to any conducting body like a metal can, then there will be a flow of charge till the potential of the negative terminal and the metal are equal because the metal body and the negative terminal are a single conducting body and a conducting body has a ...

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People have certainly measured the electron's charge and mass more than once in the last 100 years. See for example this table from the Particle Data Group, where you can find the constants you want to around 8 significant digits, much more than what was possible for Millikan. For comparison, Wikipedia claims that Millikan and Fletcher measured $e$ to be ...

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$F_1 = F_2 = F_3$. This is essentially the superposition principle. We know that in an atom, for example, in a neutral oxygen atom there are 8 protons and 8 electrons, i.e., 8 positive charged particles and 8 negative charged particles. We know it's nucleus can only carry 8 electrons around it. Now my question is why can't it carry so many electrons, ...

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Electricity is neither positive or negative as that is just a phrase attempting to explain it and is wrong ! To fully understand electricity we need to view it as a vacuum where positive is a gain in pressure and negative as like an empty void with space between them ! Here we consider space as time and the rate of charge is time over space ! Volts is the ...

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In general, no: the field of a charge distribution $\rho$ is not the same as the field of a point charge at some point therein, except for some very particular cases (the one that everyone should know is that any spherical shell of charge has an inner field of 0 but an outer field that looks exactly like all the charge is located at the center point. A ...

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Everything you say is correct in the steady state. The problem you run into is that when you remove charge from a charged capacitor to an uncharged capacitor, there is a potential difference. And somehow, you have to remove the energy from the electron that moves from one to the other. It turns out, as you calculated, that you in fact remove half of the ...

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The dipole moment of a continuous charge distribution is given by $$\mathbf{p} = \int\mathrm{d}^3\mathbf{r} \; \mathbf{r} \rho(\mathbf{r}),$$ (the moment is taken with respect to the point $\mathbf{r} = 0$). For a displaced charge distribution $\rho'(\mathbf{r}) = \rho(\mathbf{r} - \mathbf{b})$, you can use a change of integration variables to show that ...

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When the point charge is not at the center of the sphere, the electric field lines will not intersect the sphere at right angles. Consequently, there is an initial component of electric field along the surface of a conductor. We know this results in a force on the charge carriers inside the conductor, and these charge carriers will re-arrange until the ...

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Yes. The suit is of a conducting material and charge thus moves inside the suit material much more likely than through the content (the person) inside the suit.. This is called a Faraday cage and is an effective shielding mechanism from electric shocks. Excess charge will stay on the outer surface of a conducting object and charge will redistribute on the ...

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If $c$ is unknown, then you don't know it and you'll have to leave it (either directly or in another form, like $\rho$ or $Q$) as a variable in your expressions. So indeed you can either replace $c$ with $\rho/r^2$ or $Q / (4\pi R^5/3)$, but you cannot eliminate it completely unless they ask for some problem with a very fortuitous cancellation.

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Remember, the dipole is a vector. So, its not simply $p = qd$. For a general charge distribution $\rho(\mathbf r')$, you need the multipole expansion of potential in spherical coordinates, for powers of $1/r$. Meaning, take the potential, make expansion of $1/r$ powers. The $1/r$ term is the monopole. The $1/r^2$ term is the dipole. The $1/r^3$ term is the ...

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$$V = {v_r}(t) + {v_c}(t) = i(t)R + {1 \over C}\int\limits_{{t_0}}^t {i(\tau )d} \tau$$ Differentiating wrt t: $$RC{{di(t)} \over {dt}} + i(t) = 0$$ Solving differential equation: $$I(t) = {V \over R}{e^{{t \over {{\tau _0}}}}}$$ From this equation we notice initially at t = 0 $$I = {V \over R}$$ As time is increasing current starts decreasing until at ...

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By capacitor charge is meant the absolute value of the charge on each capacitor plate: $\mid Q \mid$. If the battery generates the potential difference $V$ and you connect the capacitor to the battery through a conducting wire, as shown in your picture, once the equilibrium is reached each plate of the capacitor will have a charge $Q = CV$, where $C$ is ...

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The short and simple answer is that railguns utilize a version of the Lorentz force with the term $\mathbf{j} \times \mathbf{B}$ as the main driver. They then take advantage of Faraday's law when a paramagnetic material like aluminium is exposed to a rapidly changing magnetic flux so that the $\mathbf{j} \times \mathbf{B}$-term can impart an impulse on the ...

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Background Info The first thing to do is to consider the relativistic gyrofrequency, given by: $$\Omega_{cs} = \frac{ Z_{s} \ e \ B_{o} }{ \gamma \ m_{s} }$$ where $Z_{s}$ is the charge state of species $s$, $e$ is the fundamental charge, $B_{o}$ is the quasi-static magnetic field magnitude, $\gamma$ is the relativistic Lorentz factor, and $m_{s}$ is the ...

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