Answer:
A) V = 7.5 V
B) E = 75,000 V/m
C) Q = 16.6 pC
D) V = 7.5 V
E) E = 24,000 V/m
F) Q = 52 pC
Explanation:
Given:
- The Area of plate A = ( 5 x 5 ) mm^2
- The distance between plates d = 0.10 mm
- The thickness of Mylar added t = 0.10 mm
- Voltage supplied by battery V = 7.5 V
Solution:
A) What is the capacitor's potential difference before the Mylar is inserted?
- The potential difference across the two plates is equal to the voltage provided by the battery V = 7.5 V which remains constant throughout.
B) What is the capacitor's electric field before the Mylar is inserted?
- The Electric Field E between the capacitor plates is given by:
E = V / k*d
k = 1 (air) E = 7.5 / 0.10*10^-3
E = 75,000 V/m
C) What is the capacitor's charge Q before the Mylar is inserted?
C = k*A*ε / d
k = 1 (air) C = ( 0.005^2 * 8.85*10^-12 ) / 0.0001
C = 2.213 pF
Q = C*V
Q = 7.5*(2.213)
Q = 16.6 pC
D) What is the capacitor's potential difference after the Mylar is inserted?
- The potential difference across the two plates is equal to the voltage provided by the battery V = 7.5 V which remains constant throughout.
E) What is the capacitor's electric field after the Mylar is inserted?
- The Electric Field E between the capacitor plates is given by:
E = V / k*d
k = 3.13 E = 7.5 / (3.13)0.10*10^-3
E = 24,000 V/m
F) What is the capacitor's charge after the Mylar is inserted?
C = k*A*ε / d
k = 3.13 C = 3.13*( 0.005^2 * 8.85*10^-12 ) / 0.0001
C = 6.927 pF
Q = C*V
Q = 7.5*(6.927)
Q = 52 pC
