Question

Two conductors having capacitance $ 4\mu F $ and $ 6\mu F $ are charged up to $ - 50V $ and $ 100V $ respectively, now both are connected with each other. Find common potential, final charge on both conductor, amount of charge flow and energy loss.

Answer 1

Hint :We are going to first find the equivalent capacitance for the two resistors and the total charge that gives the common potential, then the final charge on both the conductors from the voltage and capacitance values, amount of charge flow from equivalent capacitance and the voltage and energy loss from difference of final and initial energies.
The formula for the charge is given by
 $ Q = CV $
The resultant voltage is given by
 $ V' = \dfrac{{{C_1}{V_1} + {C_2}{V_2}}}{{{C_1} + {C_2}}} $
Equivalent capacitance
 $ C' = \dfrac{{{C_1}{C_2}}}{{{C_1} + {C_2}}} $
Energy loss is given by
 $ \Delta U = {U_f} - {U_i} $

Complete Step By Step Answer:
Let us first find the common potential
 The given capacitances are $ 4\mu F $ and $ 6\mu F $
The voltages up to which they are charged are $ - 50V $ and $ 100V $
Therefore, the common potential will be
 $ V' = \dfrac{{4 \times \left( { - 50} \right) + 6 \times 100}}{{4 + 6}} = 40V $
The final charge on both the conductor is
 $ {Q_1} = {C_1}V = 4 \times 40 = 150\mu C $
And
 $ {Q_2} = {C_2}V = 6 \times 40 = 240\mu C $
If the conductors are connected in series, the resulting capacitance will be
 $ C' = \dfrac{{{C_1}{C_2}}}{{{C_1} + {C_2}}} = \dfrac{{6 \times 4}}{{6 + 4}} = 2.4\mu F $
Therefore, the amount of charge flow is
 $ Q' = C'V' = 2.4 \times 40 = 96\mu C $
The energy loss can be calculated by subtracting the final energy from the initial energy
Therefore,
 $ \Delta U = \dfrac{1}{2} \cdot \dfrac{{6 \times 4}}{{6 + 4}}{\left( {100 + 50} \right)^2} \times {10^{ - 6}} = \dfrac{1}{2} \cdot \dfrac{{24}}{{10}} \times 150 \times 150 \times {10^{ - 6}} = 2700 \times {10^{ - 5}} \\
   \Rightarrow \Delta U = 2.7 \times {10^{ - 2}}J \\ $

Note :
The charging of the two conductors by supplying the voltage, makes them store the charges $ 150\mu C $ and $ 240\mu C $ respectively and also the two conductors reach a common voltage, the net flow of charge directly depends on the common voltage while the individual charges stored also depend on the common voltage, while the energy loss depends upon the initial and final voltages.