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There are no dumb questions, just bad teachers. In your case, both capacitos are in series. IT doesn't matter if they are upside or in the red boxes, it's series. How can you tell? This is a rule of thumb: you have to check if there is more than one possible path. Two elements are in series if they are one after another. There is only one path, so, if you ...


3

The circuit is drawing infinite amount of current from the battery. Have I correctly understood this ? Yes. There will be a potential difference across the resistor in parallel to capacitor and that potential difference will be resposnsible for charging it The potential across the capacitor can't change instantaneously. Therefore in the time ...


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Infinite plates have a constant electric field (at fixed charge density). Constant electric field means constant voltage gradient, so total voltage increases linearly with distance from the plate. Capacitance is charge (which is fixed) per volts (which increases with distance); hence: capacitance decreases with distance between the plates.


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Let's take it one point at a time. Capacitance is defined as the ratio of the charge(Q) of one of the two plates and the potential difference(V) across the two plates (after they have been charged, of course) Capacitance is the amount of charge that can be stored on the plates per volt across the plates, or $$C=\frac{Q}{V}$$ The statement "after ...


2

The electrolytic capacitor is the usual example of a polarized capacitor. It’s made with plates of two dissimilar metals and an electrolyte between. If voltage is applied one way, electrochemistry causes a very thin nonconducting oxide to form on one plate. That oxide acts as a very thin dielectric, so you get a high capacitance value. If applied the ...


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I'm assuming you've been asked to find the the voltage across the capacitor. Is that the question? If so, re-arrange the circuit so that R and 4R are actually a voltage divider. Same for 2R and 3R. Now you've got a capacitor across the mid-points of two voltage dividers. You can then calculate the relative voltage at each side of the cap and so get the $V_{...


1

Electric field does not give you the entire picture. You also have to consider electric potential. You are correct in observing that electric field inside the hollow sphere is zero throughout. However the electric potential is constant. And its value is given by - $$V = \frac{kQ}{R}$$ where V is the potential, R is the radius of the charged sphere and k is ...


1

it may be easier to understand the concepts of series and parallel by considering the batteries in a flashlight. They are usually on top of each other, + to -, this is in series and will deliver the voltage of both added together, usually 1.5 + 1.5 = 3 volts to the bulb. If the batteries were laying beside each other with a wire connecting - to - then to the ...


1

The two plates which have been connected are at different potentials , I'm interested to know what happens next , what changes will occur in the wire and all the plates , will a change be even there or not ? Once you disconnect the individual capacitors from their respective voltage sources they no longer have a well-defined potential. They are "floating". ...


1

You're probably familiar with the analogy in which electrical circuits can be imagined as being like water flowing in pipes. In this analogy, voltage is water pressure, current is flow-rate and so on. You can extend this to capacitors by thinking of a capacitor as a broad piece of pipe with a rubber membrane across its cross-section. In this model, as water ...


1

You are correct that the configuration here is not a parallel plate capacitor. It is a more general sort of capacitors. One in which the two plates have been "unfolded" so that their areas are actually in the same plane. Imagine a charge parallel plate capacitor with electric fields passing from one plate to the other. Now imagine tilting the two plates ...


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I think that the point that you are missing is that the ideal circuit diagram that you have drawn is an approximation of a real life situation and so you should be wary of going outside of the limits of the approximations which you have (unknowingly?) made. Consider a very simple circuit of a cell of emf $\mathcal E$ connected in series to an open switch ...


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Does the current instantly have a value of $\frac{V}{R+r}$ where V is the emf of the cell and R and r are the external and internal resistance respectively? Or does the current initially have the value $\frac{V}{R}$ and then it decreases (at a rate too fast for humans} and reaches the value $\frac{V}{R+r}$ at a steady state? If there is a substantial ...


1

So the resultant voltage across the 30 ohms resistance just at the instance of closing the switch S2 is 6V so the current just at the moment of closing the switch is $\frac{6}{30}=0.2$. This isn't a valid solution (in ideal circuit theory). First, it is typical to consider the solution just before and just after the switch is closed rather than at the ...


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For conductors all excess charge resides on the surface. Therefore, if your conductor was hollow the charge distribution would not be any different than if you had a solid cylindrical conductor. Your results will not be any different.


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What is interesting is that the OP did not include circuit diagram (3) which perhaps is a more familiar situation? In circuit 1 resistor $S$ is zero and in circuit 3 resistor $P$ is infinite and so circuits 1 and 3 are variations of circuit 2. In all three circuits the capacitor $C$ starts off uncharged and final value of the voltage across the ...


1

First of all I agree with @The Photon answer and upvoted it. I would only add the following to his first response to this statement of yours. The circuit is drawing infinite amount of current from the battery. Have I correctly understood this ? The answer is yes, but only if the battery is an ideal voltage source, that is, a source without any source ...


1

I want to address what I believe is a misconception regarding circuit A: The capacitor and resistor are connected in parallel so I think that the resistor will draw a current I=VR but the capacitor is an ideal one therefore has no resistance and therefore draws an infinite amount of current which eventually stops when the capacitor is completely ...


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