Question

Two separate capacitors, C1 and C2
C1 = 36 micro-Coulomb on 3 micro-Farad
C2 = 72 uC on X 2uF, , if zero 1
C2 had a gap of 0.2m maintained by a compressed plastic spring inside the gap, the natural spring length
was 0.5m, the compressed spring length was 0.2 m. Spring constant = 8,000 micro-Newton/ meter
Action: Connected the two capacitors in parallel
Part A
Find Q2-new, C2-new, new gap,
Hint: Capacitance has geometry parameters, build an equation with numerical coefficients
Hint: Parallel connection, V same, build an equation with numerical coefficients
Hint: F-attract equals F-spring with spring inside the gap, build an equation with numerical coefficients

Part B
Find the initial total energy, the final total energy,
Hint: use the energy formulas

Answer 1

Let us list the formulae we will need to solve this problem

1)    A is the area (we consider them to have same area), d is the distance between plates, C is the capacitance.

2) L.H.S is total capacitance for capacitors in parallel.

3) where Q is the charge on the plates and V is the potential difference between it

4)   similarly for other, where L.H.S is the force between the plates and is the charge

5) at equilibrium)

6) V between the capacitors is same after they are connected in parallel.

Use this information with the hints given in the question, to find the solution.

If both the capacitors will have same area then using (4) and (5)

q1=q2

hence this will be new q2

also using the formula (3)

we can say that

and

can be found using use the older values, it turns out to be 0.4.

Hence, substitute it.