A capacitor $C_1 = 1.0 \mu F$ is charged up to a voltage $V = 60$ V by connecting it to battery B through switch (1). Now $C_1$ is disconnected from battery and connected to a circuit consisting of two uncharged capacitors $C_2 = 3.0 \mu F$ and $C_3 = 6.0 \mu F$ through switch (2), as shown in the figure. The sum of final charges on $C_2$ and $C_3$ is :
To solve this problem, we need to calculate the total charge on the capacitors \(C_2\) and \(C_3\) after they are connected to the charged capacitor \(C_1\). Here are the steps to solve the problem:
Step 1: Calculate the initial charge on \(C_1\).
The charge \(Q_1\) on the capacitor \(C_1\) can be determined using the formula:
\(Q = CV\)
Where:
Thus,
\(Q_1 = 1.0 \times 60 = 60 \, \mu \text{C}\)
Step 2: Determine the equivalent capacitance when \(C_1\) is connected to \(C_2\) and \(C_3\).
The capacitors \(C_2\) and \(C_3\) are connected in series, so the equivalent capacitance \(C_{\text{eq}}\) is given by:
\(\frac{1}{C_{\text{eq}}} = \frac{1}{C_2} + \frac{1}{C_3}\)
Substituting the given values:
\(\frac{1}{C_{\text{eq}}} = \frac{1}{3} + \frac{1}{6} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}\)
Thus, \(C_{\text{eq}} = 2 \, \mu \text{F}\)
Step 3: Calculate the final charge distribution.
When \(C_1\) is connected to the equivalent capacitance \(C_{\text{eq}}\), the total charge is conserved. The initial charge on \(C_1\) is now distributed over \(C_1\) and the series combination \(C_2\) and \(C_3\).
The new combined capacitance becomes:
\(C_{\text{total}} = C_1 + C_{\text{eq}} = 1 + 2 = 3 \, \mu \text{F}\)
The voltage across the capacitors is:
\(V_{\text{final}} = \frac{Q_1}{C_{\text{total}}} = \frac{60}{3} = 20 \, \text{V}\)
Hence, the charge on the series combination \((Q_2 = Q_3)\) is:
\(Q_2 = C_{\text{eq}} \times V_{\text{final}} = 2 \times 20 = 40 \, \mu \text{C}\)
Conclusion: The sum of the final charges on \(C_2\) and \(C_3\) is \(40 \, \mu \text{C}\).
Using a variable frequency a.c. voltage source the maximum current measured in the given LCR circuit is 50 mA for $V = 5\sin(100t)$ The values of L and R are shown in the figure. The capacitance of the capacitor (C) used is ___________ $\mu\text{F}$. 
The space between the plates of a parallel plate capacitor of capacitance C (without any dielectric) is now filled with three dielectric slabs of dielectric constants $K_1 = 2$, $K_2 = 3$ and $K_3 = 5$ (as shown in figure). If new capacitance is $\frac{n}{3}C$ then the value of $n$ is ___________. 
Combination of four parallel plate air capacitors is shown in figure. Separation between the plate P1 & P2 is $\frac{d}{3}$ and separation between the plates P2 and P3 is d.
The equivalent capacitance of the configuration is _____________$\mu$F.
(Given $\frac{\epsilon_0 A}{d} = 4\mu$F . Where A=Area of plates)

A copper rod of mass m slides under gravity on two smooth parallel rails, with separation $l$ and set at an angle of $\theta$ with the horizontal. At the bottom, rails are joined by a resistance R. There is a uniform magnetic field B normal to the plane of the rails, as shown in the figure. The terminal speed of the copper rod is :

Using a variable frequency a.c. voltage source the maximum current measured in the given LCR circuit is 50 mA for $V = 5\sin(100t)$ The values of L and R are shown in the figure. The capacitance of the capacitor (C) used is ___________ $\mu\text{F}$. 