Q1 = C1U, Q2 = C2U
equivalent capacitor voltage U′ = U
Power Q′ = Q1 + Q2
according to the relationship between capacitor power, capacitance and voltage, we get
C′ =< table cellpadding="-1" cellspacing="-1" style="margin-right:1px">
C1U+C2U |
U |
(2) Given that the voltage of the two capacitors is U when the distance between the plates of B remains unchanged, the amount of electricity charged by both:
Q=CUCharged particle A by gravity and electric field force balance,
mg = q
U |
d |
If the distance between the plates of B is doubled, the capacitance of B becomes
C |
2 |
The equivalent capacitance of A and B in parallel is C′ = C +
C |
2 |
The total charge carried by the equivalent capacitor is Q′=2Q
So the voltage of the equivalent capacitor is ?U′=
Q′ |
C′ |
4 |
3 |
Then the electric field force F on particle A =
qU′ |
d |
4mg |
3 |
So the acceleration of particle A a=
F?mg |
m |
g |
3 |
Answer: the capacitance of the equivalent capacitor is 1.5C, the acceleration of the charged particle A
g |
3 |