# Tag Info

0

The last part is possibly the most interesting in that you have think about what happens to a dielectric when it is placed in an external electric field, or put another way; how does the movement of charges within the dielectric change the net electric field between the plates?

0

You have the right equation. Since you are not given absolute values, you should just reference your answers to the original $Q$. For example, you can say for the first part "The charge will be 0 when the voltage difference is 0. As the voltage difference increases to $\Delta V$, the charge will increase (proportionally) to $Q$". No numbers were needed... ...

0

Assuming steady state means that you have no current flowing in the branch with capacitors in it. So think of the circuit first as a potential divider made up of the battery and the two resistors. From this you can find the potential difference across the two capacitors.

0

No, the circuit is not complete. If it were complete Capacitors would discharge and 1st plate would have 0 potential. Next, potential of capacitor and plate are different. Capacitor has potential difference b/w two plates. Since both are connected in series net Capacitance = 10/3 uF . $$Q = CV = 10/3 *100 = 1000/3 uC$$ which is charge on both capacitors. ...

1

I think that the arrangement looks something like this?

0

This is a very interesting question which has at least two answers. The first answer is for an electrical/electronic engineer who wants to get a correct answer by using any systematic and coherent method. The Associated Variables Convention is used. You assign a current direction to each circuit element and then assign a plus sign at the point at which ...

0

It is possible for a capacitor to have a net charge. However, most components have a small capacitance with respect to infinity. Unless you're dealing with a huge metal ball, it will take a large voltage in order to get any appreciable charge there.

0

emf induced by inductor is negative but emf applied by battery on inductor is positive. So, that inductor tries net voltage across it is 0. Emf applied by battery is E°sinwt of which LdI/dt is applied across inductor, RI across resistance and Q/C across capacitor. -LdI/dt is induced across inductor and not applied. Now, consider direction of emf is ...

0

When you find the response of a linear circuit for a frequency $\omega$ you are taking the differential equations that describe it and looking for the homogeneous solution. The true response of the circuit is this homogeneous solution plus a particular solution which depends on the initial conditions. That is the missing term you found. The particular ...

2

Here is how: And the expression is (from wikipedia): $$2\pi \varepsilon a\sum_{n=1}^{\infty }\frac{\sinh \left( \ln \left( D+\sqrt{D^2-1}\right) \right) }{\sinh \left( n\ln \left( D+\sqrt{ D^2-1}\right) \right) }$$

1

A capacitor can store charge so if both balls were initially uncharged one could move charge from one ball to the other. Thus the balls now store charge and energy - they can be classed as a capacitor. As the separation between the balls decreases the capacitance of the system increases.

1

So the spectral approach as I remember is you make an anszats $i=\sum_k a_k f_k(t)$ or something like that, you use some theorem to say that each component of the sum is independent so you get $a_k f_k''(t) + 5 a_k f_k'(t) + 1/4 a_k f_k(t)=0$, to get the factors (characteristic polynomial is what you said I believe) My guesses where the error could be: I ...

0

I see it from a vector addition perspective. Using a single test charge in the directly between the plates, the force vectors created by individual charges on the plate begin to point more "horizontal" as the plate is moved away from the test charge. So although the individual vectors decrease in length, the horizontal components of forces from charges at ...

Top 50 recent answers are included