Two condensers $C_1$ and $C_2$ in a circuit are joined as shown in figure. The potential of point $A$ is $V_1$ and that of $B$ is $V_2$. The potential of point $D$ will be
$\frac{1}{2}\,\left( {{V_1} + {V_2}} \right)$
$\frac{{{C_2}{V_1} + {C_1}{V_2}}}{{{C_1} + {C_2}}}$
$\frac{{{C_1}{V_1} + {C_2}{V_2}}}{{{C_1} + {C_2}}}$
$\frac{{{C_2}{V_1} - {C_1}{V_2}}}{{{C_1} + {C_2}}}$
A parallel plate capacitor of capacitance $C$ is connected to a battery and is charged to a potential difference $V$ . Another capacitor of capacitance $2C$ is similarly charged to a potential difference $2V$ . The charging battery is now disconnected and the capacitors are connect in parallel to each other in such a way that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is
Two point charges $+8q$ and $-2q$ are located at $x = 0$ and $x = L$ respectively. The location of a point on the $x-$ axis at which the net electric field due to these two point charges is zero is
The equivalent capacitance between points $A$ and $B$ of the circuit shown will be
A charge $q$ is placed at the centre of the line joining two equal charges $Q$. The system of the three charges will be in equilibrium, if $q$ is equal to
A hollow cylinder has charge $q$ $C$ within it. If $\phi $ is the electric flux in unit of voltmeter associated with the curved surface $B$, the flux linked with the plane surface $A$ in unit of voltmeter will be