(2013? Bengbu three model) as shown in the figure, a, b are capacitors with capacitance C1, C2 respectively, their plates are placed horizontally, they are connected in parallel (

(1) two capacitors with equal voltage, set to U, set their power respectively Q1 and Q2

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">Q′U′ =

C1U+C2U
U
=C1+C2

(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=CU

Charged 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
= 1.5C

The total charge carried by the equivalent capacitor is Q′=2Q

So the voltage of the equivalent capacitor is ?U′=

Q′
C′
=
4
3
U

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
, direction vertically upwards.

Answer: the capacitance of the equivalent capacitor is 1.5C, the acceleration of the charged particle A

g
3