In: Physics
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
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