An electrician requires a capacitance of $6 \mu F$ in a circuit across a potential difference of $1.5 kV$. $A$ large number of $2 \mu F$ capacitors which can withstand a potential difference of not more than $500 V$ are available. The minimum number of capacitors required for the purpose is

$A$ circuit is shown in the figure for which $C_1=(3 \pm 0.011) \mu F$,$C_2=(5 \pm 0.01) \mu F$,and $C_3=(1 \pm 0.01) \mu F$. If $C$ is the equivalent capacitance across $AB$,then $C$ is given by:

Three identical capacitors $C_1, C_2$ and $C_3$ have a capacitance of $1.0 \mu F$ each and they are uncharged initially. They are connected in a circuit as shown in the figure and $C_1$ is then filled completely with a dielectric material of relative permittivity $\varepsilon_{r}$. The cell electromotive force (emf) $V_0=8 \,V$. First the switch $S_1$ is closed while the switch $S_2$ is kept open. When the capacitor $C_3$ is fully charged,$S_1$ is opened and $S_2$ is closed simultaneously. When all the capacitors reach equilibrium,the charge on $C_3$ is found to be $5 \mu C$. The value of $\varepsilon_{r}$ is:

Three isolated equal charges are placed at the three corners of an equilateral triangle as shown in the figure. The statement which is true for the net electric potential $V$ and net electric field intensity $E$ at the centre of the triangle is

In the circuit shown,$q_2$ and $q_3$ are respectively (Initially all capacitors are uncharged).

Five similar condenser plates,each of area $A$,are placed at equal distance $d$ apart and are connected to a source of $e.m.f.$ $V$ as shown in the diagram. The charge on the plates $1$ and $4$ will be: