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Question: Two conductors having capacitance \( 4\mu F \) and \( 6\mu F \) are charged up to \( - 50V \) and \(...

Two conductors having capacitance 4μF4\mu F and 6μF6\mu F are charged up to −50V- 50V and 100V100V respectively, now both are connected with each other. Find common potential, final charge on both conductor, amount of charge flow and energy loss.

Explanation

Solution

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=CVQ = CV
The resultant voltage is given by
V′=C1V1+C2V2C1+C2V' = \dfrac{{{C_1}{V_1} + {C_2}{V_2}}}{{{C_1} + {C_2}}}
Equivalent capacitance
C′=C1C2C1+C2C' = \dfrac{{{C_1}{C_2}}}{{{C_1} + {C_2}}}
Energy loss is given by
ΔU=Uf−Ui\Delta U = {U_f} - {U_i}

Complete Step By Step Answer:
Let us first find the common potential
The given capacitances are 4μF4\mu F and 6μF6\mu F
The voltages up to which they are charged are −50V- 50V and 100V100V
Therefore, the common potential will be
V′=4×(−50)+6×1004+6=40VV' = \dfrac{{4 \times \left( { - 50} \right) + 6 \times 100}}{{4 + 6}} = 40V
The final charge on both the conductor is
Q1=C1V=4×40=150μC{Q_1} = {C_1}V = 4 \times 40 = 150\mu C
And
Q2=C2V=6×40=240μC{Q_2} = {C_2}V = 6 \times 40 = 240\mu C
If the conductors are connected in series, the resulting capacitance will be
C′=C1C2C1+C2=6×46+4=2.4μFC' = \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′=2.4×40=96μCQ' = 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 150μC150\mu C and 240μC240\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.