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Capacitator problem

  • Thread starter Dell
  • Start date
  • #1
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2 parrallel boards are charged with an equal and oposite charge, when the area between them is emptied, the electric field reaches 2*105 V/m. when the area is filled with a dialectric substance, the field is reduced to 1.2*105V/m. what is the density of the induced charge??

this is what i have done so far, could someone tell me if it is correct, and if not where i am going wrong.

E=1.2*105
E0=2*105

K=E0/E=5/3

∫Kε0EdA=Qfree
==>Qfree0KEA

∫ε0EdA=Qfree+Qinduced
==>Qinduced0KEA-ε0EA

Qind/A=ε0E(1-K)

0(1.2-2)*105

=7.083*10-7C/m2
 

Answers and Replies

  • #2
674
2
∫ε0EdA=Qfree+Qinduced
==>Qinduced0KEA-ε0EA
Seems you messed up the negative sign here, but you corrected it in the following line. The rest looks good, I believe though that the final charge density should be written as negative since it makes an electric field in the opposite direction to reduce the total field. But that depends on how picky the grader is.
 

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