Question

In: Physics

Two 3.0 mm × 3.0 mm electrodes are held 0.10 mm apart and are attached to...

Two 3.0 mm × 3.0 mm electrodes are held 0.10 mm apart and are attached to 7.0 V battery. Without disconnecting the battery, a 0.10-mm-thick sheet of Mylar is inserted between the electrodes.

What is the capacitor's electric field after the Mylar is inserted?

PS: 6.98*10^4 is not correct answer

Solutions

Expert Solution

Given

   electrodes of dimensions 3mm*3mm, separated by a distance of d = 0.10 mm

   battery of 7.0 V, thickness of Mylar is t = 0.1 mm , and dielectric constant is k = 3.1


we know that Q = c*V

       V = E*d ==> Q = C*E*d
           V2/V1 = C1/C2


   and due to mylor the electric field of the capacitor decreases by k times and the capacitance increases by same amount.

Let C1 capacitance of the capacitor before introducing Mylor is


   C1 = e0*A/d
   = (8.954*10^-12**3*3*10^-6)/(0.1*10^-3) F

   = 8.0586*10^-13 F

   = 0.80586 pF


C2 is capacitance of the capacitor after introducing Mylor is


   C2 = k*e0*A/d
   = (3.1)(8.954*10^-12**3*3*10^-6)/(0.1*10^-3) F

   =2.498166*10^-12 F

   = 2.498166 pF
Q1 = C1*v = 0.80586 pF*7 = 5.64102*10^-12 C

Q2 = C2*v = 2.498166*10^-12*7 = 17.487162*10^-12 C

from the relations between V,Q,C,E
  
   E2 = C1*E1*Q2/Q1C2

   = C1*(Q1/C1*d)*Q2/Q1*C2
   = Q2/C2*d
   = (17.487162*10^-12)/(2.498166*10^-12 *0.1*10^-3) N/C

   = 70000 N/C


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