Frequency in an AC circuit given capacitance, voltage, and current

[adangerousdriver]

Homework Statement

If the voltage amplitude across an 8.50-nF capacitor is equal to 12.0 V when the current amplitude through it is 3.33 mA, the frequency is closest to:

5.20 MHz
32.6 Hz
5.2 kHz
32.6 kHz
32.6 MHz

Relevant Equations

Capacitor Reactance: Xc = 1/(ωC)
Ohm's Law: V = IXc = I/(ωC)
Frequency: f = 2πω

By combining the formula for the reactance of a capacitor with Ohm's Law for a capacitor, I can solve for angular frequency, and divide by 2π to find frequency.
The resulting equation is:

f = I/(2π VC)

Using the given values, I end up with 5.2 kHz, instead of the correct answer of 5.2 MHz. I cannot find out where I am making a mistake that throws off my order of magnitude.

 

[TSny]

I agree with your answer of 5.2 kHz.

 

[gneill]

Hi adangerousdriver,

Your calculations look correct to me. I get the same result for the given information.

Perhaps there's an error in the problem statement or the answer key. I think you should raise the question with your professor or TA?

 

[Cutter Ketch]

WOW! Ok, what the heck am I doing wrong? I’ve looked this over more than a few times, and it seems completely obvious to me that the 2 pi belongs in the numerator. However, that results in not matching ANY of the answers. I feel like I’m in an algebra twilight zone.

 

[Cutter Ketch]

Oh, I see. The equation relating frequency to angular frequency is wrong. The 2 pi should be on the other side

 

[gneill]

I feel like I’m in an algebra twilight zone.

Oh, I've definitely been there more than a few times!

It's all good now?

 

[Cutter Ketch]

Oh, I've definitely been there more than a few times!

It's all good now?

Yep. Thanks.