between the plates. 010 (part I of 6) 10.0 points A parallel-plate capacitor has
ID: 2032258 • Letter: B
Question
between the plates. 010 (part I of 6) 10.0 points A parallel-plate capacitor has a plate area of pl 163 em and a plate separation of 5.78 mm. A potential difference of 5.45 V is applied across the plates with only air between the plate +5.78 mm pla 163 em2 l 163 em2 5.45 V What is the capacitance before the dielec- tric is inserted? Answer in units of F 011 (part 2 of 6) 10.0 points The battery is then disconnected, and a piece of glass (with a dielectric constant 4.08) is inserted to completely fill the space between the plates 5.78 mm + Q11 408 4.08 163 cm 163 cm2 What is the capacitance after the dielectric is inserted? Answer in units of F 012 (part 3 of 6) 10.0 points What is the charge on the plates before the dielectric is inserted? Answer in units of CExplanation / Answer
A.
Capacitance is given by:
C = k*e0*A/d
when no dielectric inserted, k = 1
C = 8.85*10^-12*163*10^-4/(5.78*10^-3)
C = 24.96*10^-12 F
Part 2.
when dielectric inserted
C1 = k*C
C1 = 4.08*24.96*10^-12
C1 = 1.01*10^-10 F
Part 3
Q = C*V
Q = 24.96*10^-12*5.45
Q = 1.36*10^-10 C
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