This problem involves calculating the energy lost when a charged capacitor is connected to an identical uncharged capacitor.
Initial State Analysis
We start with two capacitors, each with capacitance $C = 200 \text{ pF}$. One capacitor is charged to an initial voltage $V_0 = 100 \text{ V}$.
- The initial charge $Q_0$ on the first capacitor is calculated using $Q = CV$: $Q_0 = C \times V_0 = (200 \times 10^{-12} \text{ F}) \times (100 \text{ V}) = 2 \times 10^{-8} \text{ C}$.
- The initial electrostatic energy $U_0$ stored in this capacitor is calculated using $U = \frac{1}{2}CV^2$: $U_0 = \frac{1}{2} C V_0^2 = \frac{1}{2} \times (200 \times 10^{-12} \text{ F}) \times (100 \text{ V})^2$ $U_0 = \frac{1}{2} \times 200 \times 10^{-12} \times 10000 \text{ J} = 1 \times 10^{-6} \text{ J}$.
Connecting the Capacitors
The charged capacitor is then connected to an identical uncharged capacitor. The total charge $Q_0$ is conserved and redistributed between the two capacitors.
- Since the capacitors have equal capacitance $C$, they will share the charge equally after connection. The total capacitance becomes $C_{total} = C + C = 2C$.
- The final voltage $V_f$ across both capacitors will be half the initial voltage: $V_f = \frac{V_0}{2} = \frac{100 \text{ V}}{2} = 50 \text{ V}$. Alternatively, using total charge and total capacitance: $V_f = \frac{Q_0}{C_{total}} = \frac{2 \times 10^{-8} \text{ C}}{2 \times (200 \times 10^{-12} \text{ F})} = \frac{2 \times 10^{-8}}{400 \times 10^{-12}} \text{ V} = 50 \text{ V}$.
Final Energy Calculation
The total electrostatic energy $U_f$ stored in the system of two parallel capacitors is:
- Using the total capacitance and final voltage: $U_f = \frac{1}{2} C_{total} V_f^2 = \frac{1}{2} \times (400 \times 10^{-12} \text{ F}) \times (50 \text{ V})^2$ $U_f = \frac{1}{2} \times 400 \times 10^{-12} \times 2500 \text{ J} = 500000 \times 10^{-12} \text{ J} = 0.5 \times 10^{-6} \text{ J}$.
- Alternatively, calculate energy in each capacitor: $U_f = 2 \times \left( \frac{1}{2} C V_f^2 \right) = C V_f^2 = (200 \times 10^{-12} \text{ F}) \times (50 \text{ V})^2$ $U_f = 200 \times 10^{-12} \times 2500 \text{ J} = 500000 \times 10^{-12} \text{ J} = 0.5 \times 10^{-6} \text{ J}$.
Energy Lost Calculation
The electrostatic energy lost during this process is the difference between the initial and final energies:
- $U_{lost} = U_0 - U_f$ $U_{lost} = (1 \times 10^{-6} \text{ J}) - (0.5 \times 10^{-6} \text{ J}) = 0.5 \times 10^{-6} \text{ J}$.
This energy is dissipated primarily as heat due to the current flow during charge redistribution.
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- The electric current in the circuit is given as $i = i_o(t/T)$. The r.m.s current for the period $t = 0$ to $t = T$ is ________.
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then the refractive index of the medium is ________.
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