# Homework Help: Spring and Capacitor

1. Aug 23, 2015

### Vibhor

1. The problem statement, all variables and given/known data

2. Relevant equations

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

3. 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

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2. Aug 24, 2015

### Staff: Mentor

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

3. Aug 24, 2015

### Vibhor

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?

4. Aug 24, 2015

### Staff: Mentor

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

5. Aug 24, 2015

### Vibhor

Sorry once again .

It should be (16/9)F0 .

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

6. Aug 24, 2015

### Staff: Mentor

That agrees with the answer I got.

7. Aug 24, 2015

### Vibhor

Thanks mfb .

8. Aug 24, 2015

### Vibhor

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

Right ???

9. Aug 24, 2015

### Staff: Mentor

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