Recent content by AnonBae

I figured it out! The voltage of the 1500 ohm resistor is the same amount that charged up the 100 microfarad capacitor. Thank you so much for the help :smile:

Oh. Okay For the loop specified, Each potential drop will sum to the emf or the 55 V battery. So, since 55 V = I*525 + I*1500 + I*100, since current in the same everywhere in a series loop. Thus, current would be after simplification around 0.026 from sig. figs. So V1 would be 13.65, V2 = 39...

The voltage across the branch containing the 100 μF capacitor should be 55 V, stored in the capacitor. I am not quite understanding what you mean by measuring the voltage at the open terminals. At the open terminals, the voltage would drop depending on which resistor comes before it?

The current will be flowing through the branches that do not have the capacitors, or the 525, 100, and 1500 resistors, which I did say in the original post. How will this help though in determining the charge of the capacitor if it's already fully charged? Like Q = C * V, but that does not give...

Homework Statement The circuit has been completed for several minutes. Suppose that ε = 55 V . Calculate the magnitude of charge on the 100 μF capacitor plate. Homework Equations Q = C * V The Attempt at a Solution Since the capacitors have been fully charged, current has stopped flowing at...

Homework Statement A thin semicircular rod is broken into two halves. The top half has a total charge +Q uniformly distributed along it, and the bottom half has a total charge -Q uniformly distributed along it. A negative test charge is placed at points A, B, and C. *Image attached* Consider...

a. I do know that a fractional charge of +Q contributes to the electric field at point C since no other electric field lines cross through point C besides the one horizontal to point C. I am confused as to whether that fractional charge offsets its weak electric force by the formula for the...

Homework Statement A thin semicircular rod has a total charge +Q uniformly distributed along it. A negative point charge -Q is placed as shown. (Point C is equidistant from -Q and from all points on the rod.) *Image Attached* Let Ep and Er represent the electric fields at point C due to the...