Respuesta :
Answer:
a. Q = 1881.73 x [tex]10^{-13}[/tex] C
b. As battery is not removed so, potential difference will remain same.
c. E = 21.42 x [tex]10^{3}[/tex] V/m
d. Q = 895.5 x [tex]10^{-13}[/tex] C
e. Again the potential difference will not change it will remain same as 9 V
f. E = 45 x [tex]10^{3}[/tex] V/m
Explanation:
Solution:
Here, Teflon is used so, the dielectric constant of the Teflon K = 2.1
Diameter = 2.1 cm
Radius = 2.1/2 cm
Radius = 1.05 cm
Radius = 0.015 m
Now, we need to find the area of each plate:
A = [tex]\pi r^{2}[/tex]
A = (3.14) ([tex]0.015^{2}[/tex])
A = 0.000225 [tex]m^{2}[/tex]
A = 2.25 x [tex]10^{-4}[/tex] [tex]m^{2}[/tex]
We are given the thickness of the plate which equal to the distance between the two plates.
d = 0.20 mm = 0.2 x [tex]10^{-3}[/tex] m
d = 0.2 x [tex]10^{-3}[/tex] m = distance between two plates.
Hence, the capacitance of the dielectric without the dielectric
C = [tex]\frac{E.A}{d}[/tex]
Putting up the values we get,
E = 8.85 x [tex]10^{-12}[/tex]
C = [tex]\frac{8.85 . 10^{-12} x 2.25 . 10^{-4} }{0.002}[/tex]
C = 99.5 [tex]10^{-13}[/tex]
If dielectric is included then,
[tex]C^{'}[/tex] = K C
[tex]C^{'}[/tex] = (2.1) ( 99.5 x [tex]10^{-13}[/tex])
[tex]C^{'}[/tex] = 209.08 x [tex]10^{-13}[/tex] F
As we know the voltage of the battery V = 9V So,
a) Charge before the Teflon is removed:
Q = CV
Q = [tex]C^{'}[/tex]V
Q = (209.08 x [tex]10^{-13}[/tex] F) (9V)
Q = 1881.73 x [tex]10^{-13}[/tex] C
b) Potential Difference before the Teflon is removed = ?
As battery is not removed so, potential difference will remain same.
c) Electric Field =?
As we know,
E = V/(K.d)
E = 9V/(2.1 x 0.2 x [tex]10^{-3}[/tex])
E = 21.42 x [tex]10^{3}[/tex] V/m
d) After the Teflon is removed
Q = CV
Q = (99.5 [tex]10^{-13}[/tex] ) ( 9)
Q = 895.5 x [tex]10^{-13}[/tex] C
e) Again the potential difference will not change it will remain same as 9 V
f) Electric Field = ?
E = [tex]\frac{V}{d}[/tex] (Teflon is removed)
E = 9/0.2 x [tex]10^{-3}[/tex]
E = 45 x [tex]10^{3}[/tex] V/m
