Discover the answers to your questions at Westonci.ca, where experts share their knowledge and insights with you. Get quick and reliable solutions to your questions from a community of experienced professionals on our platform. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.
Sagot :
Answer:
a) ΔV ’= 1.66 10¹ V= 16.6 V, b) U = 55.64 10⁻¹² J, c) U_f = 1.18 10⁻¹⁰ J
d) W = 6.236 10⁻¹¹ J
Explanation:
Capacitance can be found for a parallel plate capacitor
C = ε₀ [tex]\frac{A}{d}[/tex]
Let's reduce the magnitudes to the SI system
A = 9.30 cm² (1 m / 10² cm) 2 = 9.30 10⁻⁴ m²
c = 4.50 mm (1 m / 1000 mm) = 4.50 10⁻³ m
Co = 8.85 10⁻¹² 9.30 10⁻⁴ /4.50 10⁻³
Co = 1.829 10⁻¹² F
when the plates separate at d = 9.60 10⁻³ m, the capcitance changes to
C = ε₀ \frac{A}{d_1}
C = 8.85 10⁻¹² 9.30 10⁻⁴/9.60 10⁻³
C = 8.57 10⁻¹³ F
a) the potential difference
C =
since the capacitor is not discharged, let's look for the initial charge
Co = \frac{Q}{ \Delta V}
Q = C₀ ΔV
Q = 1.829 10⁻¹² 7.80
Q = 14.2662 10⁻¹² C
when the condensate plates are separated
C = \frac{Q}{ \Delta V' }
ΔV ’= Q / C
ΔV ’= 14.266 10⁻¹² / 8.57 10⁻¹³
ΔV ’= 1.66 10¹ V= 16.6 V
b) the stored energy is
U = ½ C ΔV²
for initial separation
U = ½ C₀ ΔV²
U = ½ 1.829 10⁻¹² 7.80²
U = 55.64 10⁻¹² J
c) The energy for end separation;
U_f = ½ C DV’2
U_f = ½ 8.57 10⁻¹³ 16,6²2
U_f = 1.18 10⁻¹⁰ J
d) The work
as there are no losses, the work is equal to the variation of the energy
W = ΔU = U_f -U₀
W = 1.18 10⁻¹⁰ - 55.64 10-12
W = 6.236 10⁻¹¹ J
Thank you for choosing our service. We're dedicated to providing the best answers for all your questions. Visit us again. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Thank you for trusting Westonci.ca. Don't forget to revisit us for more accurate and insightful answers.