A charged capacitor (AB) carries on its plate A a charge Q=-1.2mC
a) What is the charge carried by plate B.
b) Is the voltage U (AB)=V (A)-V (B) positive or negative.
c) The plates A and B are connected to the terminals of a resistance R: the capacitor gets discharged.
What are the charge carriers which move in the connection wires. What is the direction of the transitional electric current
a) equal and opposite
b) -charge on A, + charge on B
Vb > Va so negative
c) electrons
the electrons move through the wire from A side to B side
BUT
the current is defined as if it were positive charges moving through the wire
SO current through resistor from B side to A side
a) To find the charge carried by plate B, we need to know the total charge on the capacitor. In this case, we're given that the charge on plate A is Q = -1.2 mC. Since the total charge on a capacitor is conserved, the charge on plate B will be equal in magnitude but opposite in sign. Therefore, the charge carried by plate B will be Q = 1.2 mC.
b) The voltage difference across the capacitor, U(AB), is given by the difference in voltage between plate A (V(A)) and plate B (V(B)). Since the charge on plate A is negative, and the charge on plate B is positive (as found in part a), the voltage at plate A is lower than the voltage at plate B. Therefore, we have V(A) < V(B). As a result, the voltage difference U(AB) will be negative.
c) When the plates A and B of the capacitor are connected to the terminals of a resistance R and the capacitor begins to discharge, the charge carriers that move in the connection wires are electrons. Electrons are negatively charged particles, so they move from the negatively charged plate A (which is now at a lower voltage) towards the positively charged plate B (which is at a higher voltage).
Therefore, the transitional electric current flows from plate A to plate B, as the electrons move in the opposite direction of conventional current (which is from positive to negative).