CAPACITANCE? ...A 17.7 μF capacitor is fully charged across a 10.6 V battery....?

Q = CV

Q = (17.7 x 10^-6)(10.6) = 1.876 x 10^-4 C

This same charge is then distributed across the two capacitors...

Q = (C1 + C2)V

1.876 x 10^-4 C = (C2 + 17.7 x 10^-6)(3.59 V)

C2 = (1.876 x 10^-4 / 3.59) - 17.7 x 10^-6

C2 = 3.46 x 10^-5 F

C2 = 0.346 μF <=========================

while capacitor C is linked the definitely charged 11.0 uF capacitor, electric powered modern will flow from the 11.0 uF capacitor to capacitor C. while the present ceases, the preliminary charge of the 11.0 uF capacitor would be disbursed between the two capacitors. 11.0 uF = 11 * 10^-6 Farad Capacitance = charge ÷ Voltage charge = Capacitance * Voltage enable’s determine the charge on the 11 * 10^-6 Farad capacitor, while the voltage is eighteen volts and 3.ninety 8 volts. charge = 11 * 10^-6 * 18 = a million.ninety 8 * 10^-4 Coulomb it is the preliminary charge on the 11 * 10^-6 Farad capacitor. charge = 11 * 10^-6 * 3.ninety 8 = 4.378 * 10^-5 Coulomb it is the in simple terms right charge on the 11 * 10^-6 Farad capacitor. charge on C = a million.ninety 8 * 10^-4 – 4.378 * 10^-5 = a million.5422 * 10^-4 Coulomb Capacitance = charge ÷ Voltage = a million.5422 * 10^-4 ÷ 3.ninety 8 = 3.875 * 10^-5 F it is the capacitance of capacitor. Capacitance = ?0 * (A/d) A is the outdoors area of the plates; d is the area between the parallel plates. enable’s anticipate that the area between the parallel plates is the comparable for the two capacitors. 3.875 * 10^-5 ÷ 11 * 10^-6 = 3.5 the area of the plates in capacitor C is approximately 3.5 situations the area of the plates in the 11 * 10^-6 Farad capacitor. charge on C = a million.5422 * 10^-4 Coulomb charge on 11 * 10^-6 Farad capacitor = 4.378 * 10^-5 Coulomb a million.5422 * 10^-4 ÷ 4.378 * 10^-5 = 3.5 The charge on capacitor C is approximately 3.5 situations the charge on the 11 * 10^-6 Farad capacitor Capacitor C has extra charge, because of the fact it has extra area for the charge. To charge a capacitor, electrons are compelled from the battery to the capacitor. because all electrons have detrimental charge, electrons are continuously repelling one yet another. while the charged capacitor is linked to the uncharged capacitor, the repelling rigidity reasons electrons to flow from the charged capacitor to the uncharged capacitor. rigidity = ok * q1 * q2 ÷ d^2 in accordance to the equation above, the significance of this repelling rigidity is immediately proportional to the finished style of electron on the detrimental plate and inversely proportional to the sq. of the area between the electrons. On a extra robust plate, the electrons could nicely be farther aside. Capacitor C has extra electrons on its plates, because of the fact it has extra area for the electrons. i wish this helps you recognize this style of venture!