Two 13.0cm-diameter electrodes 0.50cm apart form a parallel-plate capacitor. The

Question

Two 13.0cm-diameter electrodes 0.50cm apart form a parallel-plate capacitor. The electrodes are attached by metal wires to the terminals of a 12V battery. After a long time, the capacitor is disconnected from the battery but is not discharged.

Found that Q = 2.8*10^-10 C correctly

Part F

What is the potential difference between the electrodes after insulating handles are used to pull the electrodes away from each other until they are 1.2cm apart? (Unit: volts)

I've tried C=e0*A/d => C=e0*(pi*radius^2)=> C=[(8.85*10^-12)(pi*0.006^2)] / 0.012 finding that C=8.341*10^-14

Then Q=CV to solve for V given the new distance separating the electrodes, => 2.8*10^-10 = (8.314*10^-14)V

finding that V=3357 volts and mastering physics is telling me this is incorrect.

What am I doing wrong?

Thanks in advance!

I will rate excellent for the correct explanation.

Explanation / Answer

Ci=e*A/d1 Cf=e*A/d2 Ci/Cf=d2/d1=1.7/0.5=3.4 --> cf=ci/3.4 V=Q/C Q - remains the same, only V will change Vf=Q/Cf=Q*3.4/Ci=3.4*Vi=12*3.4=40.8 volt

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