In an adjoining figure three capacitors $C_1,\,C_2$ and $C_3$ are joined to a battery. The correct condition will be (Symbols have their usual meanings)
$Q_1 = Q_2 = Q_3$ and $V_1 = V_2 = V_3 = V$
$Q_1 = Q_2 + Q_3$ and $V = V_1 + V_2 + V_3$
$Q_1 = Q_2 + Q_3$ and $V = V_1 + V_2$
$Q_2 = Q_3$ and $V_2 = V_3$
The equivalent capacitance of the system of capacitors between $A$ and $B$ as shown in the figure
The electric potential $(V)$ as a function of distance $(x)$ [in meters] is given by $V = (5x^2 + 10 x -9)\, Volt$. The value of electric field at $x = 1\, m$ would be......$Volt/m$
Electric field at a point varies as $r^o$ for
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
A particle of charge $Q$ and mass $m$ travels through a potential difference $V$ from rest. The final momentum of the particle is