Answer:
1(a) Between them
1(b) Negative
1(c) Unstable
1(d) 0
2(a) 12V ĀµC
2(b) 2.4V ĀµC
Explanation:
1) Both particles have a positive charge. Ā Like charges repel each other, so each particle experiences a force pushing away.
To balance the particles, there needs to be a force pushing the particles towards each other. Ā To do this, we must add a negative charge between them.
This balance is unstable; if we slightly nudge one of the particles, the particle will not return to its initial position.
The electric field at this point is 0.
2(a) I assume by "load" you mean the charge on the capacitor. Ā The voltage of the battery isn't given, so I'll leave it as V.
The charge is equal to the capacitance times the voltage.
Q = CV
Q = (12.0 ĀµF) V
Q = 12V ĀµC
2(b) Capacitors in series have an equivalent capacitance of:
1/C = 1/Cā + 1/Cā
They also have the same charge.
Q = Qā = Qā
Capacitors in parallel have an equivalent capacitance of:
C = Cā + Cā
They have the same voltage.
V = Vā = Vā
The two capacitors in series have a capacitance of:
1/C = 1/12 + 1/12
C = 6 ĀµF
This is in parallel with another capacitor, so the equivalent capacitance is:
C = 6 + 12
C = 18 ĀµF
This is in series with one more capacitor:
1/C = 1/18 + 1/12
C = 7.2 ĀµF
So the total charge in the right side of the circuit is:
Q = CV
Q = (7.2 ĀµF) V
Q = 7.2V ĀµC
That means there's 7.2V ĀµC on the bottom capacitor, and 7.2V ĀµC on the combination of capacitors on the top. Ā So the voltage across that combination is:
Q = CV
7.2V ĀµC = (18 ĀµF) V
V = 0.4V
So the charge in the left branch is:
Q = CV
Q = (6 ĀµF) (0.4V)
Q = 2.4V ĀµC
So the charge in each capacitor in that branch is also 2.4V ĀµC.
