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Electrostatic refers to the case where the fields are not time dependent. In that case the Maxwell's equations reduce to: $$\nabla \cdot E =\frac{\rho}{\epsilon_o} \\ \nabla \times E = 0 \implies E=-\nabla \phi \\ \text{then,} \nabla \cdot \nabla \phi = \nabla^2 \phi = -\frac{\rho}{\epsilon_o} $$ The solution to the last equation is: $$ \phi = ...


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The difference probably is that the graph for the gravitational potential is the one for a spherical mass distribution (or a sphere with a certain mass if you wish) and the electric one is given for a point charge. You could also draw the gravitational potential for a point mass, then it would look equivalent to your electrical potential, or the other way ...


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You say the lamp is plugged into a AC outlet, but then talk of a "wall switch". Apparently you mean that this switch controls the power to the outlet, and that a switch on the lamp is kept on, or that the lamp has no switch. If so, you should clarify this as a switched AC outlet, since most aren't. In the case of a switched AC outlet, the switch will be ...


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Basically, the system is simply that which is studied in a problem in physics. It refers to that which we want to know more about, in this case the moving electric charge in the presence of the electric field. Be cautious with the terms 'electric potential' and 'potential energy', since they're two different things. Electric potential is defined as ...


2

Observe the potential lines for a moment. You will find that for equal change in distance, there is equal change in potential. Means, if I move 0.5 m to the left, the potential increase is 10 V. In other words, we have equidistant equipotential lines which is a graphical way of denoting uniform field. Whenever you see straight equipotential lines, it means ...


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Depending on the location of the switch, the answer will change. A properly wired lamp would have no signal on the live (phase) wire, and therefore there would be no field. However, if you interrupt the neutral wire (or the switch is in the lamp, not the wall) then you will have a varying AC field because the voltage on the wire changes (and thus a small ...


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The electrons are in random motion within the cord even when it is plugged and not switched on. The motion of the electrons in this is case is random i.e., there is no preferred direction of motion of electrons or vector sum of all the thermal velocities is zero. Each electron within this conductor acts like a point source of electric filed and these micro ...


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I understand the problem now from leongz's excellent explanation and patience in the chat discussion. Thank you very much leongz for helping me out! The capacitor is being discharged through constant current. If it starts with charge Q_0, then from the definition of current we know that the charge decreases by It after time t, where I is the constant ...


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You are right that the charge gained potential energy. But this statement is only true because the charge is part of a system. We cannot talk about the electrical potential of a charge unless it is in an electric field - which means that there is "something else" that is essential for our definition of the potential. We say the "system" (the charge, plus ...


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Suppose you set the zero of potential so both conductors have zero charge at zero potential. If you then set them to potentials $V_A$ and $V_B$, you can prove that they will acquire charges \begin{align} Q_A=C_A V_A+C_{AB}V_B,\\ Q_B=C_{BA}V_A+C_B V_B, \end{align} respectively. This is the real definition of capacitance, and particularly of mutual ...



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