You invented an air capacitor. Its parallel plates are separated by 2.00 mm. Each plate carries a charge of 5.50 nC. The magnitude of the electric field of the plates is 5.50 x 105 V/m. Find the capacitance of your invented air capacitor.
Given:
"d=0.002\\:\\rm m"
"q=5.50*10^{-9}\\:\\rm C"
"E=5.50*10^5\\:\\rm V\/m"
The capacitance
"C=\\frac{q}{V}=\\frac{q}{Ed}\\\\\n=\\frac{5.50*10^{-9}}{5.50*10^{5}*0.002}=5*10^{-12}\\:\\rm F"
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