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I was wondering why, in the derivation, they set charge to Q... but if the definition is amount of charge being able to be store $Q$ is the charge separated, not stored. A charged capacitor is not electrically charged any more than a charged battery is. Rather, a charged capacitor has stored energy. To charge a parallel plate capacitor, electric ...

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By charge on capacitor, we (generally) mean the magnitude of charge on one of the plates.

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You have to be careful of your Gaussian surface. If you include both charges (from both plates) then your Gaussian surface is outside of the parallel plate capacitor and the electric field is indeed zero because there is zero net charge when encapsulating the parallel plates with your Gaussian surface. However, if you place one part side of your Gaussian ...

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your coordinates are pointing up, let's call the coordinates y-axis. at the top your initial coordinate is $y_i=h$, where $h$ is simply a distance, it's positive. at the bottom your location is $y_f=-x$, where $x$ is again a positive value, indicating how far the spring will go down. so, the equations comes up as $mgy_i=mgy_f+kx^2/2$, which is the same as ...

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It is a matter of how you define your variable. Regardless of how you define your coordinate system the result shouldn't change. Consider 2 cases, the first one being how you currently defined your vertical axis, and the second one being with an origin at the foot of the spring. Let $\Delta x$ be a positive value which is the displacement of the spring. 1st ...

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Let me try to attempt to clarify in some direction that I feel comfortable. From usual mechanics, you know the definition of work is given by : $$W = \int \bf{F}.\bf{ds}$$ for simplicity considering the work done to be in the same direction as displacement $$W = \int {F}{ds}$$ Now this is just one version of defining work. In generalising, we redefine ...

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It may just be the wording of your text that provides the confusion. Work is the same in thermodynamics as it is in mechanics. Work is an energy transfer process. In thermodynamics this often equates to energy in the forms of pressure, temperature, volume, etc. that is transformed into energy associated with position or movement (i.e. kinetic and potential ...

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This is the same idea as the direction of rotation of a ceiling fan or a box fan. Motor fans work by pulling air in from one direction, and pushing it out the other direction. Which side is the "in" side, and which is the "out" side, is determined by the direction of rotation of the blades and by the direction of the pitch of the blades. Most fans have their ...

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The electrostatic potential energy of a system of point charges is defined as the work required to be done to bring the charges constituting the system to their respective locations from infinity. Suppose we have a configuration of point charges. If the potential of the energy of the system is negative, this means work is positive. Consider two point ...

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With work you also have to take care in specifying what is doing the work. If it is the work required to move a particle or the work done by the electric field, these have different signs. If the work done by the electric field on a point charge is positive it means it is moving in the direction of the electric force acting on the point charge, therefore, $W ... 2 Rule of thumb for working it out: If you imagine letting a charge go, the direction it tends to move is toward lower potential energy. The opposite direction is toward higher potential energy. This is independent of the choice of where the zero of energy is. 1 Have to be a bit careful with potential energies, as the 0 point of potential can be arbitrarily chosen. Only changes in potential are well defined without a choice of 0 point. That said, it is often convenient to choose the 0 point at$\infty$, and this is the typical choice when talking about assembling point charges. With this choice, the potential ... 0 I believe that the reason why specific impulse has the units seconds is to prevent confusion between people who use the metric or imperial system. Because the "efficiency" of a rocket is actually indicated by the effective exhaust velocity: $$v_e=\frac{F}{\dot{m}}$$ However this has an unit of length, which is different between the metric and imperial ... 0 The Wikipedia article you link states: For all vehicles specific impulse (impulse per unit weight-on-Earth of propellant) in seconds can be defined by the following equation ... The quantity$\dot{m}g_0$is the weight flow rate of the propellant when the local gravitational acceleration is$g_0\$, so the weight-on-Earth bit of the definition implies ...

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