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I've been staring at this for 10 minutes and I'm soooo confused. Hopefully, you guys can help!

So, let's say we close the switch. From what I understand, the capacitor will charge up to $10V$ almost instantaneously, while no current will flow through the inductor. At the very beginning, there'll be $0.2A$ flowing through the resistor.

But what happens if we leave the switch closed for a long time?

My book says the current through the inductor would pick up to $0.2A$, while the current through the capacitor drops to $0A$.

That first part makes sense.

However, the book then says that after an infinite amount of time, the capacitor will be at 10 volts.

That seems wrong to me...if the current through the inductor ever stops changing, the inductor would just behave like a shorted section of a circuit, right? Wouldn't the capacitor discharge through it?

In terms of voltage drop - if the current through the resistor were ever to reach $0.2A$, that would mean that the entirety of the voltage is dropping through the resistor. There is no voltage drop through the inductor.

Buuut, since the inductor and the capacitor are in parallel, wouldn't that automatically mean there MUST be no charge on the capacitor either?

Is my book wrong in saying after an infinite amount of time with the switch closed there will be a 10-volt difference across the capacitor?

Edit:

This is an edit after reading some of your answers and comments - they've been great. I didn't elaborate enough on what my book specifically said in my original post, so this edit is just to address that. However, the fundamentals of the question remain the same.

The exact question is this:

After a very long amount of time, the switch is opened. What is the maximum current?

Then, their explanation is as follows:

We've created an $LC$ circuit, and the maximum current is $0.44A$ - Since the current through the inductor must remain continuous, the initial current through the inductor is $0.2A$. The instant the $LC$ circuit begins to oscillate, energy is stored both within the inductor and within the capacitor. We calculate the maximum current using conservation of energy:
$frac{LI_{max}^2}{2} = frac{LI_{0}^2}{2} + frac{CV_{0}^2}{2}$. Substituting in the initial values for the current, $0.2A$, and for the voltage, $10V$, we calculate the maximum current to be $0.44A$.

Like I said above, that seems wrong to me because I don't think the initial capacitor voltage would be $10V$.

In fact, I don't think the voltage across the capacitor would EVER be $10V$.

$frac{LI_{0}^2}{2} = frac{CV_{f}^2}{2}$

Plugging in an initial current of $0.2A$, I get that the maximum voltage across the capacitor is actually: $5.16 V$.

My book says the current through the inductor would pick up to 0.2A,
while the current through the capacitor drops to 0A.

This is correct. To find the DC steady state solution for this circuit, replace the inductor with a (ideal) wire and replace the capacitor with an open-circuit.

Why? In DC steady state (the solution as $trightarrowinfty$), all the circuit voltages and currents are constant.

Now, recall that the voltage across an (ideal) inductor is given by

$$v_L = Lfrac{di_L}{dt}$$

and so, since the inductor current is constant, the voltage across the inductor is zero. This is why you can replace the inductor with a wire.

For the (ideal) capacitor, the current through is given by

$$i_C = C frac{dv_C}{dt}$$

and so, since the capacitor voltage is constant, the current through the capacitor is zero. This why you can replace the capacitor with an open-circuit.

In this case, it follows that both the capacitor current and voltage are zero in DC steady state.

Is my book wrong in saying after an infinite amount of time with the
switch closed there will be a 10-volt difference across the capacitor?

Yes, if your book states that the capacitor has non-zero voltage across at infinite time, it is wrong for the reason I give above.

Tangential addendum:

From what I understand, the capacitor will charge up to 10V almost
instantaneously, while no current will flow through the inductor.

That's not correct. As the capacitor charges, the current through the inductor must increase and this inductor current means that the capacitor voltage can never reach 10V (that would require zero inductor current). This could shown by solving for the step response of the capacitor voltage which is beyond the scope of the question. However, this circuit is easily simulated with LT Spice and I've attached a plot of the capacitor voltage just after the switch is closed. See that the maximum voltage is not quite 4V.

I think your book is correct. After an infinite amount of time with the switch closed there will be no potential difference across the capacitor, there will be no current through the capacitor and there will be stationary current $0.2$ A through the inductor and the resistor. Do you see any contradictions in this picture?

Good question, as the cap is charging the inductor back emf is decreasing, back emf is 0 at infinity time. Yes the cap will certainly have 10v as long as the switch is closed. The interesting part is what happens to the charge on the cap, I believe it eventually goes to zero because the bottom of the cap will be at 10v at an infinite time.

The basic rules for ideal capacitors and inductors are as follows:

Upon Switching:

You can’t change the voltage across an ideal capacitor instantaneously.

You can’t change the current through an ideal inductor instantaneously.

After a long time with the switch closed:

A capacitor will look like an open circuit

Inductor will look like a short circuit.

Alfred Centauri has shown you the equations for the basis of these rules.

So, when you say,”…the capacitor will charge up to 10V almost instantaneously” is incorrect. But you are correct that that the current through an ideal inductor cannot change instantaneously.

Now looking at your circuit diagram it appears that just prior to closing the switch, there is a charge Q on the capacitor. Then that means the voltage across the capacitor just prior to switching is

$$V_{C}=frac{Q}{C}$$

So it appears that the inductor and capacitor are initially in parallel resonance.

Now when the switch is closed for a long time inductor is now a short-circuit with 0.2 A flowing in it and the resistor, and there is no voltage across the capacitor.

So your book is correct about the current in the inductor and wrong about the voltage across the capacitor.

Hope this helps.

From my knowledge, when you close the switch, a Back-emf will be indced across the inductor which will be equal to 10 V. Now since the inductor and capacitor are connected in parallel , the Voltage across them will always be the same ( Back emf ) . Now you can apply Kirchoff's loop law in the loop containing the battery, resistor and inductor and conclude that the magnitude of back-emf decreases with time , and hence the charge on the capacitor. After a long time , the back emf across the inductor as well as the charge on the capacitor will be zero and both will behave like short-circuited paths and the current through the resistor will be 0.2A

One way to think about this is to regard the whole right hand half of the circuit diagram - both the parallel capacitor and inductor - as a single "component" (Indeed, this is just further application of the general idea of the "lumped element model" that we are using to do these kinds of simple analyses in being able to treat a circuit as composed of isolated components at all - and it's important to point out that this model has limits, however fortunately, we are well within those in this case), with its leads being the top segment and bottom segment after the resistor, in the diagram. Then you effectively have a circuit that looks like a voltage source, followed by a resistor, followed by this strange "component", in a series loop.

As a result, the total current through that "component" can never be more than $frac{V}{R}$, or 0.2 A. Hence, looking at the sub-components of our "component" individually, i.e. the parallel inductor and capacitor, once current stops in the capacitor, the only option left is for the inductor will have to be taking that full 0.2 A as a simple conductor, as you mentioned, since current through it will no longer be changing. It will not be a "short" circuit because it is still behind that series resistor and so limited current.

With regard to your question as to whether the capacitor will discharge back through the inductor upon reaching maximum charge, the answer is no, and the reason for that is essentially the same as why a capacitor won't discharge back through a simple RC (no L) circuit: there's still a voltage drop, i.e. electric force, being applied from the voltage source and that will be holding it at maximum charge. For this case, remember that in a parallel connection, both components get the source voltage.

However, here's the interesting bit: if, on the other hand, you were to sever the wires connecting the inductor to the voltage source, then yes, indeed, the charge on the capacitor would now begin to flow through the inductor and it will discharge, as that holding force is no longer present. The inductor will then, as one may think, try to fight the resultant rising current, and it will do so on both the upswing and the downswing (as the capacitor approaches neutralization), meaning not only will it slow the capacitor's discharge, but it will at a certain point also try and "yank" charge back, and this will cause an oscillatory behavior: what you have now is a so-called "tank circuit". In theory, the charge will oscillate back and forth through the inductor forever, reversing the capacitor's polarity over and over again, however, in actuality, just as with many other conservative (i.e. not "creating energy from nothing" but rather simply keeping it in literal motion forever) "perpetual motion" schemes that seem to work in theory, the slight internal resistance in the wires due to being imperfect conductors will eventually dissipate all the energy into heat and the system will return to charge neutrality. Mathematically, the behavior of this circuit can be described as a damped harmonic oscillator, similar to a mass on a more realistic spring which is dissipating energy in internal friction due to bending the realistic material.