Question

5. A 10.0 uF parallel-plate capacitor with circular plates is connected to a 12.0 V battery. (a) What is the charge on each plate? (answer: 120 uC) (b) How much charge would be on the plates if their separation were doubled while the capacitor remained connected to the battery? (answer: 60 HC) (c) How much charge would be on the plates if the capacitor were connected to the 12.0 V battery after the radius of each plate was doubled without changing their separation? (answer: 480 HC)

Please anawer all the parts of the question

Answer 1

The capacitance of a parallel plate capacitor depends pn both the plate separation and the area ofeqch plate.

The capacitance of a parallel plate capacitor depends p n both the plate separation and the area plate. C = 10 mu f V = 12 V (a) Charge on each plate Q = C V Q = 10 times 12 mu C = 120 mu C (b) capacitance of parallel plate capacitor C = epsilon_0 A/d where A = Area of each plate d = separation b/w the plates Given that the Capacitance remain connected to the battery rightdoublearrow Voltage is same separation is doubled rightdoublearrow New Capacitance C' = epsilon_0 A/2 d = C/2 therefore Q' = C' V = 10/2 times 12 = 60 mu C (c) Given Battery remain connected d is some C' = epsilon A'/d = epsilon (pi R^2/d) Put R' = 2 R rightdoublearrow C' = epsilon_0 (pi 4 R^2)/d = 4 epsilon_0 A/d = 4 C therefore Q' = C' V = 4 times 10 times 12 = 480 mu C