Current I is flowing through the two materials having electrical conductivities o, and oy respectively
(o, > 2) as shown in the figure. The total amount of charge piled at the junction of the materials is
Answers
Explanation:
You started with two separate paths, and correctly added the capacitances in series, then got the amount of charge on each plate of each capacitor. It's the same, but backwards, along each path initially. You also know that those central segments are isolated initially, which means charges can't flow onto or off of them. You correctly showed this in your diagram, since each capacitor had 400μC, but opposite signs.
Now, the top two are connected in parallel, and basically act as one capacitor. Similarly, the bottom two are connected in parallel and act as one capacitor. You seem to have correctly added each pair's capacitance in parallel, and then added them in series to get the right individual charges on each capacitor.
The important part to remember is the sign of the charge on each capacitor. On the left side, the bottom plate of the top capacitor has -600μC, while the top plate of the bottom capacitor has +300μC, giving us a net -300μC on the left side. Similarly, the right side has net +300μC. But the part of the circuit with the switch in it is still isolated, so charge can't come from outside; it can only move from one side to the other. That's why current flowed through.
Answer:
Hello ✌️
Explanation:
You started with two separate paths, and correctly added the capacitances in series, then got the amount of charge on each plate of each capacitor. It's the same, but backwards, along each path initially. You also know that those central segments are isolated initially, which means charges can't flow onto or off of them. You correctly showed this in your diagram, since each capacitor had 400μC, but opposite signs.
Now, the top two are connected in parallel, and basically act as one capacitor. Similarly, the bottom two are connected in parallel and act as one capacitor. You seem to have correctly added each pair's capacitance in parallel, and then added them in series to get the right individual charges on each capacitor.
The important part to remember is the sign of the charge on each capacitor. On the left side, the bottom plate of the top capacitor has -600μC, while the top plate of the bottom capacitor has +300μC, giving us a net -300μC on the left side. Similarly, the right side has net +300μC. But the part of the circuit with the switch in it is still isolated, so charge can't come from outside; it can only move from one side to the other. That's why current flowed through.