My answer is off by a factor of 10^n but I can\'t figure out what cm that are se

Question

My answer is off by a factor of 10^n but I can't figure out what

cm that are separated by 0.012 mm of neoprene Problem 10: Consider a parallel plate capacitor having plates of area 1.35 (8%) rubber. You may assume the rubber has a dielectric constant 6.7. Go 50% Part (a) What is the capacitance in HF? Grade Summary 0.667 Deductions 32% Potential 68% sino Submissions cos tan (0 nt 7 8 9 110ML Attempts remaining 1 cotan asin() acos() (5% per attempt) detailed view atano acotano sinh() cosho tanh0 cotanh LND Degrees Radians 0 BACKSPACE DEL CLEAR. Feedback Submit Hints 3 for a 3% deduction. Hints remaining 0 Feedback 4 for a 4% deduction What is the capacitance of a parallel plate capacitor? How does the dielectric constant change the permittivity of the material? The permittivity in the material is simply g Keo. A 50% Part (b) What charge, 9, in coulombs, does it hold when 9.00 Vis applied to it? All content 2017 Expert TA, LLC

Explanation / Answer

here,

surafce area, a = 1.35 cm^2 = 1.35*10^-4 m^2

distance b/w the plates, d = 0.012 mm = 0.012*10^-3 m

dielectric constant, k = 6.7

capacitance, C = k*eo*A/d

capacitance, C = 6.7 * (8.85*10^-12)(1.35*10^-4)/(0.012*10^-3)

capacitance, C = 6.671*10^-10 F

capacitance, C = 6.671*10^-4 uF

charge, Q = capacitance * voltage

charge, Q = 6.671*10^-10 * 9

charge, Q = 6.004*10^-9 C

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