How Does Increasing Plate Separation Affect Capacitor Energy Storage?

AI Thread Summary
Increasing the separation between the plates of a parallel-plate capacitor reduces its capacitance, as capacitance is inversely proportional to the distance between the plates. When the capacitor is disconnected from the battery, the charge remains constant, but the voltage will change due to the altered capacitance. The energy stored in the capacitor can be recalculated using the relationship U = Q^2 / (2C), where Q is the constant charge. As the distance doubles, the capacitance halves, leading to an increase in voltage and a subsequent change in stored energy. Understanding these relationships is crucial for determining the new energy stored after the separation increase.
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Homework Statement



A parallel-plate capacitor has plates with an area of 445 cm2 and an air-filled gap between the plates that is 1.51 mm thick. The capacitor is charged by a battery to 575 V and then is disconnected from the battery. There is 4.311536113 x 10 ^-5 J of energy stored in the capacitor. The separation between the plates is now increased to 3.02 mm. How much energy is stored in the capacitor now?

Homework Equations



I know that since the capacitor is disconnected from the battery, V can change with C while Q, being trapped in the disconnected
capacitor, is constant. I calculated Q for the original
capacitor, which was about 1.5 x 10^-7 C. When the spacing, d, between the plates is increased, C changes, and V changes with it, so I know that I can't use the equation U=(1/2)CV^2 (with V=575 V) to find the energy, but I'm not sure exactly what I can use.
 
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map7s said:
...When the spacing, d, between the plates is increased, C changes, and V changes with it...

So let's see how they change.
 
Well, I thought that maybe by treating the increased space as a dielectric field and using the dielectric constant I could figure out the new values of V and C and use the Q to calculate the new Energy, but apparently I wasn't on the right track...
 
How is the capacitance related to the distance between the plates?
 
In calculating the capitance, the equation is C=(constant) x area/distance. So, b/c the battery is disconnected and the plates are pulled further away from each other, there needs to be a new equation for capitance...but I'm not sure what
 
map7s said:
...there needs to be a new equation for capitance...but I'm not sure what

Not a new equation. Only a new value for one parameter - the distance between the plates.

By how many times does the distance between the plates increase? How will this affect the capacitance? Will it become greater or smaller? By how many times?
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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