A parallel-plate capacitor with plate area3.8 cm and air-gap separation 0.25 mm
ID: 1763466 • Letter: A
Question
A parallel-plate capacitor with plate area3.8 cm and air-gap separation 0.25 mm is connected to a36 V battery, and fully charged. The battery is thendisconnected. What is the charge on the capacitor? Q= The plates are now pulled to a separation of0.45 mm. What is the charge on the capacitor now? A parallel-plate capacitor with plate area3.8 cm and air-gap separation 0.25 mm is connected to a36 V battery, and fully charged. The battery is thendisconnected. What is the charge on the capacitor? Q= The plates are now pulled to a separation of0.45 mm. What is the charge on the capacitor now? A parallel-plate capacitor with plate area3.8 cm and air-gap separation 0.25 mm is connected to a36 V battery, and fully charged. The battery is thendisconnected. What is the charge on the capacitor? Q= The plates are now pulled to a separation of0.45 mm. What is the charge on the capacitor now? What is the charge on the capacitor? Q= The plates are now pulled to a separation of0.45 mm. What is the charge on the capacitor now?Explanation / Answer
Capacitance C = A / d = [ 8.85 * 10 ^ -12 C ^ 2 / N m ^ 2 * 3.8 * 10 ^ -4 m ^ 2 ] /0.25*10^-3 m = 134.52 * 10 ^ -13 F voltage V = 36 V So, charge Q = CV = 4.842 * 10 ^ -10 Coulombs (b). Capacitance C = A / d = [ 8.85 * 10 ^ -12 C ^ 2 / N m ^ 2 * 3.8 * 10 ^ -4 m ^ 2 ] /0.45*10^-3 m = 74.73 * 10 ^ -13 F voltage V = 36 V So, charge Q = CV = 2.69* 10 ^ -10 Coulombs Capacitance C = A / d = [ 8.85 * 10 ^ -12 C ^ 2 / N m ^ 2 * 3.8 * 10 ^ -4 m ^ 2 ] /0.45*10^-3 m = 74.73 * 10 ^ -13 F voltage V = 36 V So, charge Q = CV = 2.69* 10 ^ -10 CoulombsRelated Questions
Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.