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
$\frac{1}{2} (V_1 + V_2)$
$\frac{{{C_2}{V_1} + {C_1}{V_2}}}{{{C_1} + {C_2}}}$
$\frac{{{C_1}{V_1} + {C_2}{V_2}}}{{{C_1} + {C_2}}}$
$\frac{{{C_2}{V_1} - {C_1}{V_2}}}{{{C_1} + {C_2}}}$
Four capacitors with capacitances $C_1 = 1\,μF, C_2 = 1.5\, μF, C_3 = 2.5\, μF$ and $C_4 = 0.5\, μF$ are connected as shown and are connected to a $30\, volt$ source. The potential difference between points $B$ and $A$ is....$V$
A charge $Q$ is placed at each of the opposite corners of a square. A charge $q$ is placed at each of the other two corners. If the electrical force on $Q$ is zero, then $Q/q$ equals
A charged particle with charge $q$ and mass $m$ starts with an initial kinetic energy $K$ at the centre of a uniformly charged spherical region of total charge $Q$ and radius $R$. Charges $q$ and $Q$ have opposite signs. The spherically charged region is not free to move and kinetic energy $K$ is just sufficient for the charge particle to reach boundary of the spherical charge. How much time does it take the particle to reach the boundary of the region?
Charge $q$ is uniformly distributed over a thin half ring of radius $R$. The electric field at the centre of the ring is
A network of four capacitors of capacity equal to $C_1 = C, C_2 = 2C, C_3 = 3C$ and $C_4=4C$ are conducted to a battery as shown in the figure. The ratio of the charges on $C_2$ and $C_4$ is