The plates of parallel capaitor are given charges `+4Q` and `-2Q`. The capacitor is then connected across an uncharged capacitor of same capacitances first one `(=C)`. Find the final potential difference between the plates of the first capacitor.

Each plate of a parallel plate capacitor has a charge q on it. The capacitor is now connected to a battery. Now,

Each plate of a parallel plate capacitor has a charge q on it. The capacitor is now connected to a battery. Now,

Each plate of a parallel plate capacitor has a charge q on it. The capacitor is now connected to a battery. Now,

In a parallel plate capacitor of capacitor 1muF two plates are given charges 2muC and 4muC the potential difference between the plates of the capacitor is

In a parallel plate capacitor of capacitor 1muF two plates are given charges 2muC and 4muC the potential difference between the plates of the capacitor is

Consider a parallel plate capacitor originally with a charge q_0 capacitance C_0 and potential difference Delta V_0 There is an electrostatic force of magnitude F_0 between the plates and the capacitor has a stored energy U_0 . The terminals of the capacitor are connected to another capacitor of same capacitance and charge. A dielectric slab with k_e gt 1 is inserted between the plates of the first capacitor . Which quantity decreases?

In an isolated parallel plate capacitor of capacitance C , the plates have charge Q_(1) and Q_(2) . The potential difference between the plates is

Two parallel conducting plates of a capacitor of capacitance C containing charges Q and -2Q at a distance d apart. Find out potential difference between the plates of capacitors.

Two parallel conducting plates of a capacitor of capacitance C containing charges Q and -2Q at a distance d apart. Find out potential difference between the plates of capacitors.