014 (part 5 of 6) 10.0 points What is the potential difference across the 010 (p
ID: 2031169 • Letter: 0
Question
014 (part 5 of 6) 10.0 points What is the potential difference across the 010 (part 1 of 6) 10.0 points A parallel-plate capacitor has a plate area ofplates before the dielectric is inserted? 86 cm2 and a plate separation of 4.06 m. A Answer in units of V potential difference of 6.81 V is applied across the plates with only air between the plates. 015 (part 6 of 6) 10.0 points What is the potential difference across the plates after the dielectric is inserted? 4.06 mm+ +Q Answer in units of V 6.81 V What is the e 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 6.43) is inserted to completely fill the space between the plates. 4.06 mm + 6.43 86 cm2 CIn 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 C 013 (part 4 of 6) 10.0 points What is the charge on the plates after the dielectric is inserted? Answer in units of C.Explanation / Answer
10.
given parallel plate capacitor
A = 86*10^-4 m^2
d = 4.06*10^-3 m
dV = 6.81 V
capacitance, C = A*epsilon/d
C = A*4*pi*epsilon/4*pi*d = A/4*pi*k*d
k = 8,98*10^9
hence
C = 1.877094*10^-11 F
11. after k = 6.43 dielectric is inserted
C' = kC = 1.206*10^-10 F
12. before the dielectric is inserted, charge on plates = q
q = CV = 1.278301014*10^-10 C
13. after the dielectric is inserted, since the plates are disconnected form the wires, the charge remains the same
hence q = 1.27830101*10^-10 C
14. before dielectric is inserted, PD = 6.81 V
15. after dielectric is inserted, q = C'*V
V = q/C' = q/c (1/k) = 6.81/k = 1.05909797822 V
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