Respuesta :
Answer:
A)2.53 × 10⁻¹² F
B) 350× 10⁻¹² C
C) 20,000V/m
D) 2.48 × 10⁻⁸J
E) 110.6 pF , 350 pC, 20,000V/m, 5.54 x 10⁻¹⁰ J
Explanation:
A) Given:
Area of the plates = A = 0.200 m²
Distance of separation of the plates = d = 0.007 m
Battery voltage = V = 140 V
Formula for Capacitance : C = ε₀ A/d
ε₀ = Permittivity of free space = 8.85×10⁻¹² SI units
C = [tex]\frac{(8.85\times 10^{-12})(0.2)}{0.007}[/tex]
= 2.53 × 10⁻¹² F
B) Charge on each plate: q = C V= (2.53 × 10⁻¹²)(140)
= 350 × 10⁻¹² C
C) Electric field between the plates = E = V/d
= [tex]\frac{(140)}{0.007}[/tex]
= 20000 V/m
D) Energy = 0.5 C V²
= (0.5)(2.53 × 10⁻¹²)(140)²
= 2.48 × 10⁻⁸ J
E) When the battery is disconnected, the charge and electric field remain the same.
A) Since d = 0.0016 m now,
C = [tex]\frac{(8.85\times 10^{-12})(0.2)}{0.016}[/tex]
= [tex]110.6\times 10^-12[/tex]
= 110.6 pF
B) Charge = Q =[tex]350\times 10^-12[/tex]
C) E= 20,000V
D) Energy = [tex]0.5 \frac{Q^2}{C}[/tex]
= [tex]0.5 \frac{(350\times 10^-12)^2}{(110.6\times 10^-12)}[/tex]
= [tex]5.54\times 10^{-10}[/tex] J
