The greater the applied voltage the greater will be the charge stored on the plates of the capacitor. Likewise, the smaller the applied voltage the smaller the charge. Therefore, the actual charge Q on the plates of the capacitor and can be calculated as: Where: Q (Charge, in Coulombs) = C (Capacitance, in Farads) x V (Voltage, in Volts)
As discussed earlier, the charging of a capacitor is the process of storing energy in the form electrostatic charge in the dielectric medium of the capacitor. Consider an uncharged capacitor having a capacitance of C farad. This capacitor is connected to a dc voltage source of V volts through a resistor R and a switch S as shown in Figure-1.
The capacitance of a capacitor can be defined as the ratio of the amount of maximum charge (Q) that a capacitor can store to the applied voltage (V). So the amount of charge on a capacitor can be determined using the above-mentioned formula. Capacitors charges in a predictable way, and it takes time for the capacitor to charge.
The capacitance C of a capacitor is defined as the ratio of the maximum charge Q that can be stored in a capacitor to the applied voltage V across its plates. In other words, capacitance is the largest amount of charge per volt that can be stored on the device: C = Q V
It is electrical energy storing device. In any form of capacitor, stored charge when charged by voltage V is q=cv where +cv is stored in one plate and -cv is stored in another plate. In this question charge stored should be: q = 4*10^-6 * 12 coulomb i.e q = 48 micro coulomb
The capacitors ability to store this electrical charge ( Q ) between its plates is proportional to the applied voltage, V for a capacitor of known capacitance in Farads. Note that capacitance C is ALWAYS positive and never negative. The greater the applied voltage the greater will be the charge stored on the plates of the capacitor.