Respuesta :
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
Here, we are required to evaluate parameters relating to the capacitor.
- 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
Data:
- The Area of plate A = ( 5 x 5 ) mm^2 =
- 2.5 × 10^(-5)m²
- The distance between plates d = 0.10 mm = 0.1 × 10^(-3)
- The thickness of Mylar added t = 0.10 mm = 0.1 × 10^(-3)
- Voltage supplied by battery V = 7.5 V
A) The potential difference across the two plates is equal to the voltage provided by the battery prior to the insertion of the Mylar, V = 7.5 V which remains constant throughout.
The potential difference across the two plates is equal to the voltage provided by the battery prior to the insertion of the Mylar, V = 7.5 V which remains constant throughout.B) The capacitor's Electric Field E between the capacitor plates is given as:
- The potential difference across the two plates is equal to the voltage provided by the battery prior to the insertion of the Mylar, V = 7.5 V which remains constant throughout.B) The capacitor's Electric Field E between the capacitor plates is given as: E = V / k*d
- The potential difference across the two plates is equal to the voltage provided by the battery prior to the insertion of the Mylar, V = 7.5 V which remains constant throughout.B) The capacitor's Electric Field E between the capacitor plates is given as: E = V / k*dwhere dielectric constant , k(air) = 1
- The potential difference across the two plates is equal to the voltage provided by the battery prior to the insertion of the Mylar, V = 7.5 V which remains constant throughout.B) The capacitor's Electric Field E between the capacitor plates is given as: E = V / k*dwhere dielectric constant , k(air) = 1 E = 7.5 /(1 × 0.10*10^-3)
- E = 75,000 V/m
C) The capacitor's charge Q before the Mylar is inserted can be evaluated as follows;
- C = k × A × ε /d
ε /d
- ε /d C =
ε /d C = (1×2.5×10^(-5)×8.85*10^-12)/0.0001
- ε /d C = (1×2.5×10^(-5)×8.85*10^-12)/0.0001 C = 2.213 × 10^(-12) = 2.213 pF
- ε /d C = (1×2.5×10^(-5)×8.85*10^-12)/0.0001 C = 2.213 × 10^(-12) = 2.213 pF Q = C × V
- Q = 7.5*(2.213)
- Q = 16.6 pC
D) The potential difference across the two plates, after the mylar has been inserted, V = 7.5 V which remains constant throughout.
E) The Electric Field, E between the capacitor plates is given by:
where the dielectric constant of the mylar = 3.13
- E = V / k × d
- E = 7.5 / (3.13)0.10*10^-3
- E = 24,000 V/m
F) C = k×A×ε / d
- C = 3.13×( 0.005^2 × 8.85×10^-12 ) / 0.0001
- C = 6.927pC
- Q = C × V
- Q = 7.5 × (6.927)
- Q = 52 pC
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