Question
The magnitude of the electric field between two plates of a parallel plate capacitor is 2.4×10^5 N/C. Each plate carries a charge whose magnitude is o.15 mC. What is the area of each plate?
Answers
bobpursley
E= Q/area * 1/epsilon
okay so I did that, but I still did not end up with the correct answer. The book says that the area of each plate should be 7.1×10^-2 m^2, but I keep getting an answer that is completely off from that.
bobpursley
area= Q/Ee= .15E-3/(2.4E5*8.85E-12)=71m^2
Is the mC milliCoulombs, or microCoulombs?
It microCoulombs, then area= .071m^2, which agrees with your text.
Is the mC milliCoulombs, or microCoulombs?
It microCoulombs, then area= .071m^2, which agrees with your text.
Thank you so much. I've got an AP test tomorrow and that really helps :)
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