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If you insert a conductor without touching either plates you end up with two capacitors in series with the widths less than that of the original capacitor you had before you inserted the conductor.


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I would suggest finding the electric field due to the three rods and integrating that to find the potential difference. That way you would not need a reference point, since all you are finding is a potential difference. Keep in mind that the wires have a diameter $a$, so that when integrating the electric field, you should approach each wire no closer than ...


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It depends. The time evolution stays the same: $V(t)=V_0 e^{-5}e^{-t/\tau}$, the "non stabilized" part of the total voltage is just smaller. Often this part is considered constant for practical purposes.


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After $5 \tau$, the voltage across the capacitor is about $.7$% of what it was originally. There are definitely situations where this $.7$% could be significant, so when the problem says you can consider this voltage to be zero, it probably means that the accuracy of any tool you would use to measure the system is low enough that it wouldn't be able to ...


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Since many metals form a thin oxide layer of relatively high resistivity, they can be used to affect the dielectric properties of a capacitor made from that material. For example, aluminium, tantalum and even copper all form oxide layers with high resistivity (compared to the respective metals) and high relative permittivity (dielectric constant). Since ...


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Force per unit length would be given by $$F=\alpha E.$$ For an infinite line charge the electric field at a distance $d$ is, by Gauss' Law, $$E=\frac{\alpha}{2\pi \epsilon_{0} d}.$$ The dielectric is made of dipoles, so you should be able to figure out why it makes no difference to the field outside the wire. And the two wires are far enough apart that we ...


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First point is that you would normally get more charge at the edges. Second point - if you increase the gap then the capacitance would be reduce and the quantity of charge that could be stored for a particular potential difference would be reduced. Remember electric field is charge / distance so decreasing the gap would increase the electric field You ...


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Let's do some calculus. Suppose you have two plates, almost parallel (off by an angle $\alpha$). The plates lie in the XY plane, from $(0, 0)$ to $(x_1, y_1)$. At $x = 0$, the plates are separated by a distance $z_0$, and at $x = x_1$, the plates are separated by a distance $z_1$. We'll now consider an infinitesimally small element of both plates. (Since ...


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It helps to think about Maxwell's equations here. The first one of which is $$ \nabla \cdot \mathbf{E} = \rho / \epsilon_{0} $$ where $\rho$ is the charge density. Electrostatics is always like a chicken and egg problem: charges make fields, but fields move charges around, so it can lead to some confusion. It turns out that in the steady state ...



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