Two identical capacitors, have the same capacitance $C$. One of them is charged to potential ${V_1}$ and the other to ${V_2}$. The negative ends of the capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is
$\frac{1}{4}C(V_1^2 - V_2^2)$
$\frac{1}{4}C(V_1^2 + V_2^2)$
$\frac{1}{4}C{\left( {{V_1} - {V_2}} \right)^2}$
$\frac{1}{4}C{\left( {{V_1} + {V_2}} \right)^2}$
In a uniform electric field, a cube of side $1\ cm$ is placed. The total energy stored in the cube is $8.85\ μJ$ . The electric field is parallel to four of the faces of the cube. The electric flux through any one of the remaining two faces is
A $5\, \mu F$ capacitor is charged fully by a $220\,V$ supply. It is then disconnected from the supply and is connected in series to another uncharged $2.5\;\mu F$ capacitor. If the energy change during the charge redistribution is $\frac{ X }{100} \;J$ then value of $X$ to the nearest integer is$.....$
A capacitor $4\,\mu F$ charged to $50\, V$ is connected to another capacitor of $2\,\mu F$ charged to $100 \,V$ with plates of like charges connected together. The total energy before and after connection in multiples of $({10^{ - 2}}\,J)$ is
A $4\, \,\mu F$ condenser is charged to $400\, V$ and then its plates are joined through a resistance. The heat produced in the resistance is.......$J$
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.......$g$ [All the elements other than plates are massless and nonconducting] :-