For an inverting voltage follower, C1 would be part of an LP filter. But in the depicted non-inverting voltage follower circuit, what is the purpose of C1 (if any)?
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The purpose of the resistor R2 is to eliminate the DC offset caused by the op-amp input bias currents. If the bias currents are exactly matched, then the voltage drop across each 1K resistor (R1 and R2) is the same and the output voltage is equal to the input voltage.
Fine, but if the op-amp has significant input capacitance or there is a lot of stray capacitance on the traces then the feedback signal has a lag and the AC output may be higher than the input. In an extreme case (very high resistance and an op-amp with a lot of input capacitance), it could oscillate. Putting the capacitor across the resistor deals with that.
1K is such a low value that this would typically not be a problem in practice.
This Answer has been edited (see below the original answer) because it has received down votes and was closed off. I requested that it be re-opened because I think the down-voters could not understand what I was trying to say.
If OA1 is assumed to be an ideal op-amp then R1, R2 and C1 serve no purpose at all. You might just as well short the resistors and leave the cap off. Then again, V1 is shown as a voltage source so the op-amp is totally redundant too - you might as well connect the sine wave voltage source to whatever connects to the op-amp's output: -
I am still saying the addition of the capacitor has no practical purpose and this may call into question Spehro's answer because the reasons I give contradict it. Consider a simple inverting op-amp configuration: -
Then ask yourself the question, does this work (i.e. not oscillate) without adding a feedback capacitor thus: -
The answer is "YES".
In other words it doesn't oscillate because of parasitic input capacitance to ground unless of course it is a totally badly designed PCB. If the question asked was related to the inverting configuration (why a capacitor was placed across R2) the great and the good would say that it was acting as a low pass filter.
I asked Spehro what he thought and he said "I did not see an error in your post" so, given he is a top contributor to this site I can only conclude that some amount of bad-thinking has been going on.
So, down-voters, please do consider what I'm saying and try looking a little bit further past the end of your noses.
And, if you think I'm being rude or insulting, that's the price for me taking the time to ressurect this answer from the grave in order to give the readers of stack-exchange a little more to think about.