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Calculating current

  • #1

Homework Statement



A glass tube with a cross-sectional area of 8^-2 m^2 contains an ionizd gas. The densities are 10^15 positive ions/m^3 and 10^11 free electrons/m3. Under the influence of an applied voltage, the postive ions are moving with an average axial velocity of 6 x 10^3 m/s. At that same point the axial velocity of the electrons is 1000 times as great. Calculate the electric current moving through.

Homework Equations



A= c/s = dq(+ ions)/dt + dq(- ions)

e = 1.6602E-19

The Attempt at a Solution



10^15*6x10^3*.08=4.8e17
4.8e17 * 1.602x10^-19 = .0769c/s
10^11*6x10^6*.08
4.8e16 * 1.602x10^-19 = .00769c/s

.00769+.0769= .0845c/s = 84.5mA


This isn't correct, but i don't know why? please help.
 

Answers and Replies

  • #2
393
0

Homework Statement



A glass tube with a cross-sectional area of 8^-2 m^2 contains an ionizd gas. The densities are 10^15 positive ions/m^3 and 10^11 free electrons/m3. Under the influence of an applied voltage, the postive ions are moving with an average axial velocity of 6 x 10^3 m/s. At that same point the axial velocity of the electrons is 1000 times as great. Calculate the electric current moving through.

Homework Equations



A= c/s = dq(+ ions)/dt + dq(- ions)

e = 1.6602E-19

The Attempt at a Solution



10^15*6x10^3*.08=4.8e17
4.8e17 * 1.602x10^-19 = .0769c/s
10^11*6x10^6*.08
4.8e16 * 1.602x10^-19 = .00769c/s

.00769+.0769= .0845c/s = 84.5mA


This isn't correct, but i don't know why? please help.
If you wrote it right, you're using .08 which is 8 x 10^-2. The problem says 8^-2 = 1/64
 
  • #3
thanks, you are correct about that mistake with the .08 thing. i fixed that, but there is still something wrong with my method. any thoughts?
 

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