Spring & Capacitor Homework: Equations & Solutions

[Vibhor]

Homework Statement

Homework Equations

Force of attraction between the plates = (1/2)(QV)/d

The Attempt at a Solution

Initial charge on the capacitor Q₁=CV
Final charge on the capacitor Q₂=2CV

When switch is open force between the plates F₁ = (1/2)(Q₁V)/d₁ = (1/2)(CV²)/d₁

When switch is closed force between the plates F₂ = (1/2)(Q₂V)/d₂ = (1/2)(2CV²)/d₂

d₂ = (3/2)d₁

F₁/F₂ = 3/4 or F₂ = (4/3)F₁

The spring force also becomes 4/3 times of the initial force i.e (4/3)F₀

Is it correct ?

Many Thanks

 

[mfb]

inal charge on the capacitor Q₂=2CV

Be careful here, the capacitance changes.
The potential changes as well, which influences the calculation of F₂.

 

[Vibhor]

Be careful here, the capacitance changes.
The potential changes as well, which influences the calculation of F₂.

F₁/F₂ = 8/9 or F₂ = (9/8)F₁

The spring force also becomes 9/8 times of the initial force i.e (9/8)F₀

Is it correct now?

 

[mfb]

Can you show your steps? I get a different result.

 

[Vibhor]

Sorry once again .

It should be (16/9)F₀ .

If it is wrong , i will surely show you the steps .

 

[mfb]

That agrees with the answer I got.

 

[Vibhor]

Thanks mfb .

 

[Vibhor]

The spring stretches in both the situations . More in the latter case , as attractive force between the plates gets stronger.

Right ?

 

[mfb]

Right. Which also means the system has to be moved a bit, otherwise the solution doesn't make sense.