dry air will break down if the electric field exceeds 3.0*106 V/m. If the electr
ID: 2030948 • Letter: D
Question
dry air will break down if the electric field exceeds 3.0*106 V/m. If the electric field exceeds this value, a current can form, generating a spark. What amount of charge can be placed on a circular-plate capacitor before reaching this maximum electric field strength if the diameter of each plate is 6.0 cm? Assuming nothing else changes, what would be the charge if a piece of Quartz were inserted into the capacitor (Dielectric constant = 4.3)?
ALLEquation(s) to use
Final Solution:
TABLE 17-3 Dielectric Constants (at 20 C) Dielectrie Dielecrie TABLE 18-1 Resistivity and Temperature Coefficients lat 20 C constant strength (V/m) Air (1 atm 006 3x 10 159 x 10 1.68 104 22 1010Copper × 1 244 10 Polystyrene 2624 X Vinyl(plastic) 2-4 50xUP Aluminum 265 x o 971 x 10 98 x 10 5.6 ×104 3.7 1510 Iran 8x0 12×1of Mercury 14 x 10 Nkchnome (Ni,Fe, Cr alloy) Glass, Pyrex 5 Rubber, ncoprene 67 100 10 360 x 10 01-10 Carhon (graphite 5 x10Geum 150x10boktors (1-300px 103 Water (liquid) so Hard rubber 0-10 alc depend strongy on the presee el eve lightd inpities CHAPTER 18 Electric Currents CHAPTER 19 DC Circuits CHAPTER 20 Magnetism ioNI 2 r ,40 = 4 * 10-7Explanation / Answer
We know that
Q = C*V
C = e0*A/d
V = E*d
Q = (e0*A/d)*(E*d)
Q = e0*A*E = e0*pi*d^2*E/4
Using given values:
Q = 8.85*10^-12*pi*0.06^2*3*10^6/4
Q = 7.51*10^-8 C
Part B.
If quartz is inserted, then
C = k*e0*A/d
then
Q = k*e0*pi*d^2*E/4
Using given values:
Q = 4.3*8.85*10^-12*pi*0.06^2*3*10^6/4
Q = 3.23*10^-7 C
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