Am I overthinking this problem?
A 15uF capacitor is connected ot a 55V battery and becomes fully charged. The battery is removed and the circuit is left open. A slab of dielectric material is inserted to completely fill the space between the plates. It has a dielectric constant of 4.8...
Alrighty based on the equation:
\frac{kQq}{0.55 - x} = E_{k} + \frac{kQq}{0.55}
I found that x is equal to 0.32m, which is the distance that the particle moves before stopping.
Does this look a little better? This number seems more appropriate to me.
So I am really confused now.
The equation should read:
\frac{kQq}{0.55} - \frac{kQq}{x} = E_{k}?
And Ek is \frac{kQq}{0.55} + 1/2mv^2?
Therefore \frac{kQq}{0.55} - \frac{kQq}{x} = \frac{kQq}{0.55} + 1/2mv^2 ?
So I plugged in the numbers based on my final equation and got 1.05m. So this means that the fired particle goes past the fixed charge before it would turn back, does this make sense?
I'm not sure, so if
\frac{kQq}{0.55 - x} = E_{k}
then Ek is the PE + KE of q?
So my equation would be:
\frac{kQq}{0.55 - x} = \frac{kQq}{0.55} + 1/2mv^2?
So here's what I've come up with:
Q = fixed charge
q = moving charge
kQq/(x-0.55m) = kQq/r +(1/2mv^2)
Solving for x.
I'm not sure if x-0.55 is correct? Or even if I've set up the equation correctly?
Another problem I'm struggling with this one I don't have a clue where to start. Any hints to a starting point would be greatly appreciated.
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