How Does Inserting a Dielectric Change Capacitor Voltage?

AI Thread Summary
Inserting a dielectric slab into a charged capacitor reduces the voltage across its plates due to the increase in capacitance. The initial voltage was 11.7 V, and with a dielectric constant (κ) of 3.1, the new voltage can be calculated as V = 11.7 V / 3.1. The calculations indicate that the expected voltage should be approximately 3.77 V, but previous attempts yielded incorrect results. The discussion suggests that the homework program may require rounding to a specific number of significant figures, possibly leading to a final answer of 3.8 V. Understanding the relationship between charge, capacitance, and voltage is crucial for solving this problem accurately.
arod2812
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Homework Statement


An empty capacitor is connected to a 11.7-V battery and charged up. The capacitor is then disconnected from the battery, and a slab of dielectric material (κ = 3.1) is inserted between the plates.
Find the magnitude of the amount by which the potential difference across the plates changes.


Homework Equations


I did the following steps but did not get the right answer. Is there another step to the problem?

The Attempt at a Solution


q=CV, therefore C=q/V
C is affected by the k... therefore C(dielectric)= k*C
then the equation reads: q/V = k (q/11.7)... and the Q's cancel out.

The final equation I worked with is (1/V)=3.1 (1/11.7). The answer I got however is not correct. help.
 
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Looks like you are on the right track. Since Q is constant (no place for the charge to go), CV must be constant. Since C goes up by 3.1x, V must do the inverse. What answer exactly did you try?
 
the answer i tried was 3.77... originally i had tried 36.27 which was wrong. But it turns out that both are wrong. What do you think?
 
Well, that's pretty much what I get for 11.7/3.1. And unless I'm missing something, that seems like the right way to get it -- even talks about using the dielectric insertion method for measuring k here:

http://en.wikipedia.org/wiki/High-k

How many sig figs does this homework program expect? Maybe they just want 3.8V?
 
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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