(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 (

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′ =

(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:

Charged particle A by gravity and electric field force balance,

mg = q

If the distance between the plates of B is doubled, the capacitance of B becomes

The equivalent capacitance of A and B in parallel is C′ = C +

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

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

Then the electric field force F on particle A =

So the acceleration of particle A a=

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