The lower plate of a parallel plate capacitor is supported on a rigid rod. The upper plate is suspended from one end of a balance. The two plates are joined together by a thin wire and subsequently disconnected. The balance is then counterpoised. Now a voltage V = 5000 volt is applied between the plates. The distance between the plates is d = 5 mm and the area of each plate is A `=100 cm^(2)`. Then find out the additional mass placed to maintain balance. [All the elements other than plates are massless and nonconducting]

Answer

`(in_(0))/(2g)xx10^(10) kg = 4.425 g`